CBSE Class 10 · Maths
CBSE Class 10 Maths Some Applications of Trigonometry Competency Based Questions
Help your child master CBSE Class 10 Maths Some Applications of Trigonometry with these expert-verified competency based questions aligned to the latest board exam pattern. Each question comes with a full step-by-step solution to build your child’s confidence in solving real-world height and distance problems. Prepared by Angle Belearn’s CBSE specialists.
CBSE Class 10 Maths Some Applications of Trigonometry — Questions with Solutions
If the height of the tower is equal to the length of its shadow, then the angle of elevation of the Sun is:
Solution
(b) According to the condition, the height of the tower is equal to the length of its shadow, i.e.,
$$ BC = AB \quad \dots (1) $$.png)
Let $\theta$ be the angle of elevation of the Sun. Now in right-angled $\triangle ABC$:
$$ \tan \theta = \frac{BC}{AB} \quad \text{(from eq. (1))} $$ $$ \Rightarrow \tan \theta = 1 $$ $$ \Rightarrow \tan \theta = \tan 45^\circ $$ $$ \Rightarrow \theta = 45^\circ $$
$$ BC = AB \quad \dots (1) $$
Let $\theta$ be the angle of elevation of the Sun. Now in right-angled $\triangle ABC$:$$ \tan \theta = \frac{BC}{AB} \quad \text{(from eq. (1))} $$ $$ \Rightarrow \tan \theta = 1 $$ $$ \Rightarrow \tan \theta = \tan 45^\circ $$ $$ \Rightarrow \theta = 45^\circ $$
The angle subtended by a tower of height 200 meters at a point 200 meters from the base is:
Solution
(b) Let the angle subtended by a tower (RQ) at a point $P$ from the base $Q$ is $\theta$. Given, $RQ = 200\,\text{m}$, $PQ = 200\,\text{m}$ and $\angle RPQ = \theta$.
.png)
Now, in right-angled $\triangle PQR$:
$$ \tan \angle RPQ = \frac{RQ}{PQ} $$ $$ \Rightarrow \tan \theta = \frac{200}{200} = 1 $$ $$ \Rightarrow \tan \theta = \tan 45^\circ $$ $$ \Rightarrow \theta = 45^\circ $$
Now, in right-angled $\triangle PQR$:$$ \tan \angle RPQ = \frac{RQ}{PQ} $$ $$ \Rightarrow \tan \theta = \frac{200}{200} = 1 $$ $$ \Rightarrow \tan \theta = \tan 45^\circ $$ $$ \Rightarrow \theta = 45^\circ $$
If a pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground, then Sun’s elevation is:
Solution
(a) Let $PQ = 6\,\text{m}$ be the height of the pole and $RQ = 2\sqrt{3}\,\text{m}$ be its shadow.
.png)
Let the angle of elevation of the Sun be $\theta$. In right-angled $\triangle PQR$:
$$ \tan \theta = \frac{PQ}{RQ} $$ $$ \Rightarrow \tan \theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} $$ $$ \Rightarrow \tan \theta = \tan 60^\circ $$ $$ \Rightarrow \theta = 60^\circ $$
Let the angle of elevation of the Sun be $\theta$. In right-angled $\triangle PQR$:$$ \tan \theta = \frac{PQ}{RQ} $$ $$ \Rightarrow \tan \theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} $$ $$ \Rightarrow \tan \theta = \tan 60^\circ $$ $$ \Rightarrow \theta = 60^\circ $$
The angle subtended by a vertical pole of height 100 m at a point on the ground $100\sqrt{3}$ m from the base is:
Solution
(d) Given, height of the pole, $AB = 100\,\text{m}$ and distance of a point $P$ from the base $A$, $AP = 100\sqrt{3}\,\text{m}$.
.png)
Let $\theta$ be the angle subtended by pole ($AB$) to the point $P$. In right-angled $\triangle PAB$:
$$ \tan \theta = \frac{AB}{PA} $$ $$ \Rightarrow \tan \theta = \frac{100}{100\sqrt{3}} = \frac{1}{\sqrt{3}} $$ $$ \Rightarrow \tan \theta = \tan 30^\circ $$ $$ \Rightarrow \theta = 30^\circ $$
Let $\theta$ be the angle subtended by pole ($AB$) to the point $P$. In right-angled $\triangle PAB$:$$ \tan \theta = \frac{AB}{PA} $$ $$ \Rightarrow \tan \theta = \frac{100}{100\sqrt{3}} = \frac{1}{\sqrt{3}} $$ $$ \Rightarrow \tan \theta = \tan 30^\circ $$ $$ \Rightarrow \theta = 30^\circ $$
The ratio of the length of a rod and its shadow is $1 : \sqrt{3}$, then the angle of elevation of the Sun is:
Solution
(b) Let $C$ be the position of the Sun. Let $BC$ and $AB$ be the length of rod and length of the shadow.
.png)
Given: $\dfrac{\text{Length of rod}}{\text{Length of shadow}} = \dfrac{1}{\sqrt{3}} = \dfrac{BC}{AB} \quad \dots (1)$
In right-angled $\triangle ABC$:
$$ \tan \theta = \frac{BC}{AB} $$ $$ \Rightarrow \tan \theta = \frac{1}{\sqrt{3}} $$ $$ \Rightarrow \tan \theta = \tan 30^\circ $$ $$ \Rightarrow \theta = 30^\circ $$
Given: $\dfrac{\text{Length of rod}}{\text{Length of shadow}} = \dfrac{1}{\sqrt{3}} = \dfrac{BC}{AB} \quad \dots (1)$In right-angled $\triangle ABC$:
$$ \tan \theta = \frac{BC}{AB} $$ $$ \Rightarrow \tan \theta = \frac{1}{\sqrt{3}} $$ $$ \Rightarrow \tan \theta = \tan 30^\circ $$ $$ \Rightarrow \theta = 30^\circ $$
From a point $P$ on a level ground, the angle of elevation of the top of a tower is $30^\circ$. If the tower is 100 m high, the distance of point $P$ from the foot of the tower is:
Solution
(c) Let $QR = 100\,\text{m}$ be the height of the tower and point $P$ makes an angle of elevation of $30^\circ$, i.e., $\angle PQR = 30^\circ$.
.png)
In right-angled $\triangle PQR$:
$$ \tan 30^\circ = \frac{RQ}{PQ} $$ $$ \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{PQ} $$ $$ \Rightarrow PQ = 100\sqrt{3}\,\text{m} $$ $$ \Rightarrow PQ = 100 \times 1.73 = 173\,\text{m} $$
In right-angled $\triangle PQR$:$$ \tan 30^\circ = \frac{RQ}{PQ} $$ $$ \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{PQ} $$ $$ \Rightarrow PQ = 100\sqrt{3}\,\text{m} $$ $$ \Rightarrow PQ = 100 \times 1.73 = 173\,\text{m} $$
The angle of elevation of the top of the tower is $60^\circ$ and the horizontal distance from the observer’s eye to the foot of the tower is 100 m, then the height of the tower will be:
Solution
(c) Let $BC = h$ metre be the height of the tower and distance from the observer to the foot of the tower be $AB = 100\,\text{m}$.
.png)
In right-angled $\triangle ABC$:
$$ \tan 60^\circ = \frac{BC}{AB} $$ $$ \Rightarrow \sqrt{3} = \frac{h}{100} $$ $$ \Rightarrow h = 100\sqrt{3} $$
In right-angled $\triangle ABC$:$$ \tan 60^\circ = \frac{BC}{AB} $$ $$ \Rightarrow \sqrt{3} = \frac{h}{100} $$ $$ \Rightarrow h = 100\sqrt{3} $$
The string of a kite in the air is 50 m long and it makes an angle of $60^\circ$ with the horizontal. Assuming the string to be straight, the height of the kite from the ground is:
Solution
(d) Let the string ($PK$) of a kite K in air is $PK = 50\,\text{m}$ long and it makes an angle $KPB = 60^\circ$ with the horizontal $PB$.
.png)
Suppose the height of the kite from the ground ($KB$) = $h\,\text{m}$. Now, in right-angled $\triangle PBK$:
$$ \sin 60^\circ = \frac{KB}{PK} $$ $$ \Rightarrow \frac{\sqrt{3}}{2} = \frac{h}{50} \Rightarrow h = \frac{50\sqrt{3}}{2} = 25\sqrt{3}\,\text{m} $$ So, the required height is $25\sqrt{3}$ m.
Suppose the height of the kite from the ground ($KB$) = $h\,\text{m}$. Now, in right-angled $\triangle PBK$:$$ \sin 60^\circ = \frac{KB}{PK} $$ $$ \Rightarrow \frac{\sqrt{3}}{2} = \frac{h}{50} \Rightarrow h = \frac{50\sqrt{3}}{2} = 25\sqrt{3}\,\text{m} $$ So, the required height is $25\sqrt{3}$ m.
The angle of elevation of a ladder leaning against a wall is $60^\circ$ and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:
Solution
(d) Let $BC$ be the height of the wall and $AC = l$ be the length of the ladder leaning against a wall.
The ladder $AC$ makes an angle of elevation of $60^\circ$, i.e., $\angle CAB = 60^\circ$. Let $AB = 4.6$ m.
In right-angled $\triangle ABC$:
$$ \cos 60^\circ = \frac{AB}{AC} $$ $$ \Rightarrow \frac{1}{2} = \frac{4.6}{l} $$ $$ \Rightarrow l = 9.2\,\text{m} $$
The ladder $AC$ makes an angle of elevation of $60^\circ$, i.e., $\angle CAB = 60^\circ$. Let $AB = 4.6$ m.In right-angled $\triangle ABC$:
$$ \cos 60^\circ = \frac{AB}{AC} $$ $$ \Rightarrow \frac{1}{2} = \frac{4.6}{l} $$ $$ \Rightarrow l = 9.2\,\text{m} $$
A boy standing on top of a tower of height 20 m observes that the angle of depression of a car on the road is $60^\circ$. The distance between the foot of the tower and the car must be:
[Use $\sqrt{3} = 1.73$]
[Use $\sqrt{3} = 1.73$]
Solution
(b) Let $BC = 20$ m be the height of the tower. Let $A$ be the position of the car and $C$ be the position of the boy.
At point $C$, the boy makes an angle of depression of $60^\circ$, i.e., $\angle HCA = 60^\circ$.
Here, $\angle BAC = \angle HCA = 60^\circ$ (alternate angles) in right-angled $\triangle ABC$:
$$ \tan 60^\circ = \frac{BC}{AB} $$ $$ \Rightarrow \sqrt{3} = \frac{20}{AB} $$ $$ \Rightarrow AB = \frac{20}{\sqrt{3}} = \frac{20}{3} \times \sqrt{3} \times 1.73 = 6.67 \times 1.73 = 11.54\,\text{m} $$
At point $C$, the boy makes an angle of depression of $60^\circ$, i.e., $\angle HCA = 60^\circ$.Here, $\angle BAC = \angle HCA = 60^\circ$ (alternate angles) in right-angled $\triangle ABC$:
$$ \tan 60^\circ = \frac{BC}{AB} $$ $$ \Rightarrow \sqrt{3} = \frac{20}{AB} $$ $$ \Rightarrow AB = \frac{20}{\sqrt{3}} = \frac{20}{3} \times \sqrt{3} \times 1.73 = 6.67 \times 1.73 = 11.54\,\text{m} $$
There is a fire incident in the house. The house door is locked so, the fireman is trying to enter the house from the window. He places the ladder against the wall such that its top reaches the window as shown in the figure.
Based on the above information, solve the following questions:
Q1. If the window is 6 m above the ground and the angle made by the foot of the ladder to the ground is 30°, then the length of the ladder is:
(a) 8 m (b) 10 m (c) 12 m (d) 14 m
Q2. If the fireman places the ladder 5 m away from the wall and the angle of elevation is observed to be 30°, then the length of the ladder is:
(a) 5 m (b) $\dfrac{10}{\sqrt{3}}$ m (c) $\dfrac{15}{\sqrt{2}}$ m (d) 20 m
Q3. If the fireman places the ladder 2.5 m away from the wall and the angle of elevation is observed to be 60°, then find the height of the window. (Take $\sqrt{3} = 1.73$)
(a) 4.325 m (b) 5.5 m (c) 6.3 m (d) 2.5 m
Q4. If the height of the window is 8 m above the ground and the angle of elevation is observed to be 45°, then the horizontal distance between the foot of the ladder and the wall is:
(a) 2 m (b) 4 m (c) 6 m (d) 8 m
Q5. If the fireman gets a 9 m long ladder and the window is at 6 m height, then how far should the ladder be placed?
(a) 5 m (b) $3\sqrt{5}$ m (c) 3 m (d) 4 m
Based on the above information, solve the following questions:Q1. If the window is 6 m above the ground and the angle made by the foot of the ladder to the ground is 30°, then the length of the ladder is:
(a) 8 m (b) 10 m (c) 12 m (d) 14 m
Q2. If the fireman places the ladder 5 m away from the wall and the angle of elevation is observed to be 30°, then the length of the ladder is:
(a) 5 m (b) $\dfrac{10}{\sqrt{3}}$ m (c) $\dfrac{15}{\sqrt{2}}$ m (d) 20 m
Q3. If the fireman places the ladder 2.5 m away from the wall and the angle of elevation is observed to be 60°, then find the height of the window. (Take $\sqrt{3} = 1.73$)
(a) 4.325 m (b) 5.5 m (c) 6.3 m (d) 2.5 m
Q4. If the height of the window is 8 m above the ground and the angle of elevation is observed to be 45°, then the horizontal distance between the foot of the ladder and the wall is:
(a) 2 m (b) 4 m (c) 6 m (d) 8 m
Q5. If the fireman gets a 9 m long ladder and the window is at 6 m height, then how far should the ladder be placed?
(a) 5 m (b) $3\sqrt{5}$ m (c) 3 m (d) 4 m
Answer
1. Let $AC$ be the length of the ladder in right-angled $\triangle ABC$.
$$ \sin 30^\circ = \frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{6}{AC} \Rightarrow AC = 12\,\text{m} $$
So, option (c) is correct.
2. In right-angled $\triangle ABC$, $\cos 30^\circ = \dfrac{AB}{AC}$.
$$ \frac{\sqrt{3}}{2} = \frac{5}{AC} \Rightarrow AC = \frac{10}{\sqrt{3}}\,\text{m} $$
So, option (b) is correct.
3. Let $BC$ be the height of the window from the ground in right-angled $\triangle ABC$.
$$ \tan 60^\circ = \frac{BC}{AB} \Rightarrow \frac{\sqrt{3}}{1} = \frac{BC}{2.5} \Rightarrow BC = 2.5 \times 1.73 = 4.325\,\text{m} $$
So, option (a) is correct.
4. Let $AB$ be the horizontal distance between the foot of the ladder and wall in right-angled $\triangle ABC$.
$$ \tan 45^\circ = \frac{BC}{AB} \Rightarrow 1 = \frac{8}{AB} \Rightarrow AB = 8\,\text{m} $$
So, option (d) is correct.
5. Let the required distance be $x$ in right-angled $\triangle ABC$.
$$ (9)^2 = x^2 + (6)^2 \quad \text{(By Pythagoras theorem)} $$
$$ \Rightarrow 81 – 36 = x^2 \Rightarrow 45 = x^2 \Rightarrow x = \sqrt{3} \times \sqrt{5}\,\text{m} = 3\sqrt{5}\,\text{m} $$
So, option (b) is correct.
$$ \sin 30^\circ = \frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{6}{AC} \Rightarrow AC = 12\,\text{m} $$
So, option (c) is correct.2. In right-angled $\triangle ABC$, $\cos 30^\circ = \dfrac{AB}{AC}$.
$$ \frac{\sqrt{3}}{2} = \frac{5}{AC} \Rightarrow AC = \frac{10}{\sqrt{3}}\,\text{m} $$
So, option (b) is correct.3. Let $BC$ be the height of the window from the ground in right-angled $\triangle ABC$.
$$ \tan 60^\circ = \frac{BC}{AB} \Rightarrow \frac{\sqrt{3}}{1} = \frac{BC}{2.5} \Rightarrow BC = 2.5 \times 1.73 = 4.325\,\text{m} $$
So, option (a) is correct.4. Let $AB$ be the horizontal distance between the foot of the ladder and wall in right-angled $\triangle ABC$.
$$ \tan 45^\circ = \frac{BC}{AB} \Rightarrow 1 = \frac{8}{AB} \Rightarrow AB = 8\,\text{m} $$
So, option (d) is correct.5. Let the required distance be $x$ in right-angled $\triangle ABC$.
$$ (9)^2 = x^2 + (6)^2 \quad \text{(By Pythagoras theorem)} $$
$$ \Rightarrow 81 – 36 = x^2 \Rightarrow 45 = x^2 \Rightarrow x = \sqrt{3} \times \sqrt{5}\,\text{m} = 3\sqrt{5}\,\text{m} $$
So, option (b) is correct. A group of students of class-X visited India Gate on an education trip. The teacher narrated that India Gate, official name is Delhi Memorial, originally called All-India War Memorial, is a monumental sandstone arch in New Delhi dedicated to the troops of British India who died in wars fought between 1914 and 1919. India Gate, located at the eastern end of the Rajpath, is about 138 feet (42 meters) in height.
Based on the given information, solve the following questions:
Q1. What is the angle of elevation, if they are standing at a distance of $42\sqrt{3}$ m away from the monument?
(a) $0^\circ$ (b) $30^\circ$ (c) $45^\circ$ (d) $60^\circ$
Q2. They want to see the tower (monument) at an angle of $60^\circ$. So, they want to know the distance where they should stand and hence find the distance. [Use $\sqrt{3} = 1.732$]
(a) 24.24 m (b) 20.12 m (c) 42 m (d) 25.64 m
Q3. If the altitude of the Sun is at $30^\circ$, then the height of the vertical tower that will cast a shadow of length 30 m is:
(a) $10\sqrt{3}$ m (b) $\dfrac{10}{\sqrt{3}}$ m (c) $\dfrac{20}{\sqrt{3}}$ m (d) $20\sqrt{3}$ m
Q4. The ratio of the length of a rod and its shadow is $24 : 8\sqrt{3}$. The angle of elevation of the Sun is:
(a) $30^\circ$ (b) $60^\circ$ (c) $45^\circ$ (d) $90^\circ$
Q5. The angle formed by the line of sight with the horizontal when the object viewed is above the horizontal level, is:
(a) angle of elevation (b) angle of depression (c) corresponding angle (d) complete angle
Based on the given information, solve the following questions:Q1. What is the angle of elevation, if they are standing at a distance of $42\sqrt{3}$ m away from the monument?
(a) $0^\circ$ (b) $30^\circ$ (c) $45^\circ$ (d) $60^\circ$
Q2. They want to see the tower (monument) at an angle of $60^\circ$. So, they want to know the distance where they should stand and hence find the distance. [Use $\sqrt{3} = 1.732$]
(a) 24.24 m (b) 20.12 m (c) 42 m (d) 25.64 m
Q3. If the altitude of the Sun is at $30^\circ$, then the height of the vertical tower that will cast a shadow of length 30 m is:
(a) $10\sqrt{3}$ m (b) $\dfrac{10}{\sqrt{3}}$ m (c) $\dfrac{20}{\sqrt{3}}$ m (d) $20\sqrt{3}$ m
Q4. The ratio of the length of a rod and its shadow is $24 : 8\sqrt{3}$. The angle of elevation of the Sun is:
(a) $30^\circ$ (b) $60^\circ$ (c) $45^\circ$ (d) $90^\circ$
Q5. The angle formed by the line of sight with the horizontal when the object viewed is above the horizontal level, is:
(a) angle of elevation (b) angle of depression (c) corresponding angle (d) complete angle
Answer
1. Let the angle of elevation be $\theta$. Height of the monument $(BC) = 42$ m and $AB = 42\sqrt{3}$ m.
Now, in right-angled $\triangle ABC$:
$$ \tan \theta = \frac{BC}{AB} = \frac{42}{42\sqrt{3}} \Rightarrow \tan \theta = \frac{1}{\sqrt{3}} = \tan 30^\circ \Rightarrow \theta = 30^\circ $$ So, option (b) is correct.
2. Let the required distance be $x$ m. Given, $\theta = 60^\circ$ and height $(BC) = 42$ m.
$$ \tan 60^\circ = \frac{42}{x} \Rightarrow \sqrt{3} = \frac{42}{x} \Rightarrow x = \frac{42}{\sqrt{3}} = \frac{42\sqrt{3}}{3} = 14\sqrt{3} \Rightarrow 14 \times 1.732 = 24.24\,\text{m} $$
So, option (a) is correct.
3. Let the height of the vertical tower be $h$ m. Given $\theta = 30^\circ$ and shadow length $(AB) = 30$ m.
$$ \tan 30^\circ = \frac{h}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} = 10\sqrt{3}\,\text{m} $$
So, option (a) is correct.
4. Let $AC = 24k$ and $BC = 8\sqrt{3}k$.
In right-angled $\triangle ACB$:
$$ \tan \theta = \frac{AC}{BC} = \frac{24k}{8\sqrt{3}k} = \frac{3}{\sqrt{3}} \Rightarrow \tan \theta = \tan 60^\circ \Rightarrow \theta = 60^\circ $$ So, option (b) is correct.
5. The angle formed by the line of sight with the horizontal when the object viewed is above the horizontal level, is angle of elevation.
So, option (a) is correct.
Now, in right-angled $\triangle ABC$:$$ \tan \theta = \frac{BC}{AB} = \frac{42}{42\sqrt{3}} \Rightarrow \tan \theta = \frac{1}{\sqrt{3}} = \tan 30^\circ \Rightarrow \theta = 30^\circ $$ So, option (b) is correct.
2. Let the required distance be $x$ m. Given, $\theta = 60^\circ$ and height $(BC) = 42$ m.
$$ \tan 60^\circ = \frac{42}{x} \Rightarrow \sqrt{3} = \frac{42}{x} \Rightarrow x = \frac{42}{\sqrt{3}} = \frac{42\sqrt{3}}{3} = 14\sqrt{3} \Rightarrow 14 \times 1.732 = 24.24\,\text{m} $$
So, option (a) is correct.3. Let the height of the vertical tower be $h$ m. Given $\theta = 30^\circ$ and shadow length $(AB) = 30$ m.
$$ \tan 30^\circ = \frac{h}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} = 10\sqrt{3}\,\text{m} $$
So, option (a) is correct.4. Let $AC = 24k$ and $BC = 8\sqrt{3}k$.
In right-angled $\triangle ACB$:$$ \tan \theta = \frac{AC}{BC} = \frac{24k}{8\sqrt{3}k} = \frac{3}{\sqrt{3}} \Rightarrow \tan \theta = \tan 60^\circ \Rightarrow \theta = 60^\circ $$ So, option (b) is correct.
5. The angle formed by the line of sight with the horizontal when the object viewed is above the horizontal level, is angle of elevation.
So, option (a) is correct.
Suppose a straight vertical tree is broken at some point due to storm and the broken part is inclined at a certain distance from the foot of the tree.
Based on the above information, solve the following questions:
Q1. If the top of the upper part of the broken tree touches the ground at a distance of 30 m (from the foot of the tree) and makes an angle of inclination $30^\circ$, then find the height of the remaining part of the tree.
Q2. Find the height of the straight vertical tree.
Q3. If the height of a tree is 6 m, which is broken by wind in such a way that its top touches the ground and makes an angle $30^\circ$ with the ground. Find the length of the broken part of the tree.
OR
If $AB = 10\sqrt{3}$ m and $AD = 2\sqrt{3}$ m, then find $CD$. In right-angled triangle $\triangle ABC$, $\angle ABC = 60^\circ$.
Based on the above information, solve the following questions:Q1. If the top of the upper part of the broken tree touches the ground at a distance of 30 m (from the foot of the tree) and makes an angle of inclination $30^\circ$, then find the height of the remaining part of the tree.
Q2. Find the height of the straight vertical tree.
Q3. If the height of a tree is 6 m, which is broken by wind in such a way that its top touches the ground and makes an angle $30^\circ$ with the ground. Find the length of the broken part of the tree.
OR
If $AB = 10\sqrt{3}$ m and $AD = 2\sqrt{3}$ m, then find $CD$. In right-angled triangle $\triangle ABC$, $\angle ABC = 60^\circ$.
Answer
1. Let $AB$ be the tree of height $h$ m and let it be broken at a height of $x$ m as shown in the figure.
Clearly, $CD = AC = (h – x)$ m. Now, in right-angled triangle $\triangle CBD$:
$$ \tan 30^\circ = \frac{BC}{BD} = \frac{x}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{30} \Rightarrow x = \frac{30}{\sqrt{3}} = 10\sqrt{3}\,\text{m} $$ Thus, the height of the remaining part of the tree is $10\sqrt{3}$ m.
2. In right-angled $\triangle CBD$:
$$ \cos 30^\circ = \frac{DB}{DC} = \frac{30}{DC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{30}{DC} \Rightarrow DC = \frac{60}{\sqrt{3}} = 20\sqrt{3}\,\text{m} $$ From part (1), $BC = x = 10\sqrt{3}$ m. Thus, the height of the straight vertical tree is:
$$ AB = DC + BC = 20\sqrt{3} + 10\sqrt{3} = 30\sqrt{3}\,\text{m} $$
3. Here, $h = 6$ m and $\theta = 30^\circ$, so $DC = AC = (6 – x)$ m.
$$ \sin 30^\circ = \frac{BC}{CD} \Rightarrow \frac{1}{2} = \frac{x}{6 – x} \Rightarrow 6 – x = 2x \Rightarrow 3x = 6 \Rightarrow x = 2 $$
So, the broken part of the tree, $AC = 6 – x = 6 – 2 = 4$ m.
Or: Clearly, $BD = AB – AD = (10\sqrt{3} – 2\sqrt{3}) = 8\sqrt{3}$ m.
In right-angled $\triangle BCD$:
$$ \sin 60^\circ = \frac{BD}{DC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8\sqrt{3}}{DC} \Rightarrow DC = 16\,\text{m} $$
Clearly, $CD = AC = (h – x)$ m. Now, in right-angled triangle $\triangle CBD$:$$ \tan 30^\circ = \frac{BC}{BD} = \frac{x}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{30} \Rightarrow x = \frac{30}{\sqrt{3}} = 10\sqrt{3}\,\text{m} $$ Thus, the height of the remaining part of the tree is $10\sqrt{3}$ m.
2. In right-angled $\triangle CBD$:
$$ \cos 30^\circ = \frac{DB}{DC} = \frac{30}{DC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{30}{DC} \Rightarrow DC = \frac{60}{\sqrt{3}} = 20\sqrt{3}\,\text{m} $$ From part (1), $BC = x = 10\sqrt{3}$ m. Thus, the height of the straight vertical tree is:
$$ AB = DC + BC = 20\sqrt{3} + 10\sqrt{3} = 30\sqrt{3}\,\text{m} $$
3. Here, $h = 6$ m and $\theta = 30^\circ$, so $DC = AC = (6 – x)$ m.
$$ \sin 30^\circ = \frac{BC}{CD} \Rightarrow \frac{1}{2} = \frac{x}{6 – x} \Rightarrow 6 – x = 2x \Rightarrow 3x = 6 \Rightarrow x = 2 $$
So, the broken part of the tree, $AC = 6 – x = 6 – 2 = 4$ m.Or: Clearly, $BD = AB – AD = (10\sqrt{3} – 2\sqrt{3}) = 8\sqrt{3}$ m.
In right-angled $\triangle BCD$:
$$ \sin 60^\circ = \frac{BD}{DC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8\sqrt{3}}{DC} \Rightarrow DC = 16\,\text{m} $$
Sandeep and his sister Dolly visited their uncle’s place — Bir, Himachal Pradesh. During daytime, Sandeep, who is standing on the ground, spots a paraglider at a distance of $24\,\text{m}$ from him at an elevation of $30^\circ$. His sister Dolly is also standing on the roof of a $6\,\text{m}$ high building and observes the elevation of the same paraglider at $45^\circ$. Sandeep and Dolly are on opposite sides of the paraglider.
1. Find the distance of the paraglider from the ground.
2. Find the value of $PD$.
3. Find the distance between the paraglider and Dolly.
OR Find the distance between Sandeep and the base of the building.
1. Find the distance of the paraglider from the ground.2. Find the value of $PD$.
3. Find the distance between the paraglider and Dolly.
OR Find the distance between Sandeep and the base of the building.
Answer
1. In the right-angled $\triangle AQD$, we have:
$$ \sin 30^\circ = \frac{DQ}{AD} \Rightarrow \frac{1}{2} = \frac{DQ}{24} $$ Thus, the distance of the paraglider from the ground is $DQ = 12\,\text{m}$.
2. We have $PQ = BC = 6\,\text{m}$. Now, as $DQ = 12\,\text{m}$:
$$ DP = DQ – PQ = 12 – 6 = 6\,\text{m} $$
3. In right-angled $\triangle BPD$, we have:
$$ \sin 45^\circ = \frac{DP}{BD} \Rightarrow \frac{1}{\sqrt{2}} = \frac{6}{BD} \Rightarrow BD = 6\sqrt{2}\,\text{m} $$ Thus, the distance of the paraglider from Dolly is $6\sqrt{2}$ m.
Or: In right-angled $\triangle AQD$: $\cos 30^\circ = \dfrac{AQ}{AD} \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AQ}{24} \Rightarrow AQ = 12\sqrt{3}$ m.
In right-angled $\triangle BPD$: $\cos 45^\circ = \dfrac{BP}{BD} \Rightarrow \dfrac{1}{\sqrt{2}} = \dfrac{BP}{6\sqrt{2}} \Rightarrow BP = 6$ m.
Thus, the distance between Sandeep and the base of the building is:
$$ AQ + BP = 12\sqrt{3} + 6 = 6(2\sqrt{3} + 1)\,\text{m} $$
$$ \sin 30^\circ = \frac{DQ}{AD} \Rightarrow \frac{1}{2} = \frac{DQ}{24} $$ Thus, the distance of the paraglider from the ground is $DQ = 12\,\text{m}$.
2. We have $PQ = BC = 6\,\text{m}$. Now, as $DQ = 12\,\text{m}$:
$$ DP = DQ – PQ = 12 – 6 = 6\,\text{m} $$
3. In right-angled $\triangle BPD$, we have:
$$ \sin 45^\circ = \frac{DP}{BD} \Rightarrow \frac{1}{\sqrt{2}} = \frac{6}{BD} \Rightarrow BD = 6\sqrt{2}\,\text{m} $$ Thus, the distance of the paraglider from Dolly is $6\sqrt{2}$ m.
Or: In right-angled $\triangle AQD$: $\cos 30^\circ = \dfrac{AQ}{AD} \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AQ}{24} \Rightarrow AQ = 12\sqrt{3}$ m.
In right-angled $\triangle BPD$: $\cos 45^\circ = \dfrac{BP}{BD} \Rightarrow \dfrac{1}{\sqrt{2}} = \dfrac{BP}{6\sqrt{2}} \Rightarrow BP = 6$ m.
Thus, the distance between Sandeep and the base of the building is:
$$ AQ + BP = 12\sqrt{3} + 6 = 6(2\sqrt{3} + 1)\,\text{m} $$
Radio towers are used for transmitting a range of communication services including radio and television. On a similar concept, a radio station tower was built in two sections A and B. The tower is supported by wires from a point O.
Distance between the base of the tower and point O is $36\,\text{cm}$. From point O, the angle of elevation of the top of Section B is $30^\circ$ and the angle of elevation of the top of Section A is $45^\circ$.
Q1. Find the length of the wire from the point O to the top of Section B.
Q2. Find the height of the Section A from the base of the tower.
Q3. Find the distance AB. Or Find the area of $\triangle OPB$.
Distance between the base of the tower and point O is $36\,\text{cm}$. From point O, the angle of elevation of the top of Section B is $30^\circ$ and the angle of elevation of the top of Section A is $45^\circ$.Q1. Find the length of the wire from the point O to the top of Section B.
Q2. Find the height of the Section A from the base of the tower.
Q3. Find the distance AB. Or Find the area of $\triangle OPB$.
Answer
1. Let the length of the wire from point O to the top of section B, i.e., $OB = l$ m. Given, $OP = 36$ cm and $\angle BOP = 30^\circ$.
Now in right-angled $\triangle BPO$:
$$ \cos 30^\circ = \frac{OP}{OB} \Rightarrow \frac{\sqrt{3}}{2} = \frac{36}{OB} \Rightarrow l = 72 \times \frac{\sqrt{3}}{3} = 24\sqrt{3}\,\text{cm} $$
2. Given, $\angle AOP = 45^\circ$ and $OP = 36$ cm. Now in right-angled $\triangle AOP$:
$$ \tan 45^\circ = \frac{AP}{OP} \Rightarrow 1 = \frac{AP}{36} \Rightarrow AP = 36\,\text{cm} $$
Thus, the height of Section A from the base of the tower is $AP = 36$ cm.
3. In right-angled $\triangle BPO$:
$$ \tan 30^\circ = \frac{BP}{OP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BP}{36} \Rightarrow BP = 12\sqrt{3}\,\text{cm} $$ $$ AB = AP – BP = 36 – 12\sqrt{3} = 12(3 – \sqrt{3})\,\text{cm} $$
Or: Since $\triangle BPO$ is a right-angled triangle:
$$ \text{Area of } \triangle BPO = \frac{1}{2} \times OP \times BP = \frac{1}{2} \times 36 \times 12\sqrt{3} = 216\sqrt{3}\,\text{cm}^2 $$
Now in right-angled $\triangle BPO$:$$ \cos 30^\circ = \frac{OP}{OB} \Rightarrow \frac{\sqrt{3}}{2} = \frac{36}{OB} \Rightarrow l = 72 \times \frac{\sqrt{3}}{3} = 24\sqrt{3}\,\text{cm} $$
2. Given, $\angle AOP = 45^\circ$ and $OP = 36$ cm. Now in right-angled $\triangle AOP$:
$$ \tan 45^\circ = \frac{AP}{OP} \Rightarrow 1 = \frac{AP}{36} \Rightarrow AP = 36\,\text{cm} $$
Thus, the height of Section A from the base of the tower is $AP = 36$ cm.3. In right-angled $\triangle BPO$:
$$ \tan 30^\circ = \frac{BP}{OP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BP}{36} \Rightarrow BP = 12\sqrt{3}\,\text{cm} $$ $$ AB = AP – BP = 36 – 12\sqrt{3} = 12(3 – \sqrt{3})\,\text{cm} $$
Or: Since $\triangle BPO$ is a right-angled triangle:
$$ \text{Area of } \triangle BPO = \frac{1}{2} \times OP \times BP = \frac{1}{2} \times 36 \times 12\sqrt{3} = 216\sqrt{3}\,\text{cm}^2 $$
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the length of the shadow of a vertical pole is equal to its height, then the angle of elevation of the Sun is $45^\circ$.
Reason (R): Trigonometric ratio, tangent is defined as
$$ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} $$
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the length of the shadow of a vertical pole is equal to its height, then the angle of elevation of the Sun is $45^\circ$.
Reason (R): Trigonometric ratio, tangent is defined as
$$ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} $$
Answer
Correct Option: (A) — Both A and R are true and R is the correct explanation of A.
Since the length of the shadow equals the height of the pole:
$$ \tan \theta = \frac{\text{Height of the pole}}{\text{Length of shadow}} = \frac{h}{h} = 1 = \tan 45^\circ $$ Therefore, $\theta = 45^\circ$. The reason correctly explains the assertion.
Since the length of the shadow equals the height of the pole:
$$ \tan \theta = \frac{\text{Height of the pole}}{\text{Length of shadow}} = \frac{h}{h} = 1 = \tan 45^\circ $$ Therefore, $\theta = 45^\circ$. The reason correctly explains the assertion.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The angle of elevation of the top of a tower is $60^\circ$. If the height of the tower and its base is tripled then angle of elevation of its top will also be tripled.
Reason (R): In an equilateral triangle of side $3\sqrt{3}$ cm, the length of the altitude is $4.5$ cm.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The angle of elevation of the top of a tower is $60^\circ$. If the height of the tower and its base is tripled then angle of elevation of its top will also be tripled.
Reason (R): In an equilateral triangle of side $3\sqrt{3}$ cm, the length of the altitude is $4.5$ cm.
Answer
Correct Option: (D) — A is false but R is true.
The assertion is incorrect. If the height and base of a tower are tripled, the ratio of height to base remains the same, so $\tan\theta$ does not change and the angle of elevation remains $60^\circ$, not $180^\circ$. Angles do not scale linearly.
The reason is correct. In an equilateral triangle:
$$ \text{Altitude} = \frac{\sqrt{3}}{2} \times \text{Side} = \frac{\sqrt{3}}{2} \times 3\sqrt{3} = \frac{3 \times 3}{2} = 4.5\,\text{cm} $$
The assertion is incorrect. If the height and base of a tower are tripled, the ratio of height to base remains the same, so $\tan\theta$ does not change and the angle of elevation remains $60^\circ$, not $180^\circ$. Angles do not scale linearly.
The reason is correct. In an equilateral triangle:
$$ \text{Altitude} = \frac{\sqrt{3}}{2} \times \text{Side} = \frac{\sqrt{3}}{2} \times 3\sqrt{3} = \frac{3 \times 3}{2} = 4.5\,\text{cm} $$
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): Suppose a bird was sitting on a tree. A person was sitting on the ground and saw the bird, which makes an angle such that $\tan \theta = \dfrac{12}{5}$. The distance from bird to the person is $13$ units.
Reason (R): In a right-angled triangle,
$$ (\text{Hypotenuse})^2 = (\text{Side})^2 + (\text{Base})^2 $$
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): Suppose a bird was sitting on a tree. A person was sitting on the ground and saw the bird, which makes an angle such that $\tan \theta = \dfrac{12}{5}$. The distance from bird to the person is $13$ units.
Reason (R): In a right-angled triangle,
$$ (\text{Hypotenuse})^2 = (\text{Side})^2 + (\text{Base})^2 $$
Answer
Correct Option: (A) — Both A and R are true and R is the correct explanation of A.
The assertion is correct. Given $\tan \theta = \dfrac{12}{5}$, opposite = 12 and adjacent = 5. Using Pythagoras theorem:
$$ (\text{Hypotenuse})^2 = 12^2 + 5^2 = 144 + 25 = 169 \Rightarrow \text{Hypotenuse} = \sqrt{169} = 13\,\text{units} $$ The reason is correct and directly supports the assertion.
The assertion is correct. Given $\tan \theta = \dfrac{12}{5}$, opposite = 12 and adjacent = 5. Using Pythagoras theorem:
$$ (\text{Hypotenuse})^2 = 12^2 + 5^2 = 144 + 25 = 169 \Rightarrow \text{Hypotenuse} = \sqrt{169} = 13\,\text{units} $$ The reason is correct and directly supports the assertion.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The angle of elevation of the top of the tower is $30^\circ$ and the horizontal distance from the observer’s eye to the foot of the tower is $50$ m, then the height of the tower will be $\dfrac{50}{3}\sqrt{3}$.
Reason (R): While using the concept of angle of elevation/depression, the triangle should be a right-angled triangle.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The angle of elevation of the top of the tower is $30^\circ$ and the horizontal distance from the observer’s eye to the foot of the tower is $50$ m, then the height of the tower will be $\dfrac{50}{3}\sqrt{3}$.
Reason (R): While using the concept of angle of elevation/depression, the triangle should be a right-angled triangle.
Answer
Correct Option: (A) — Both A and R are true and R is the correct explanation of A.
The assertion is correct. Using $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$:
$$ \frac{1}{\sqrt{3}} = \frac{h}{50} \Rightarrow h = \frac{50}{\sqrt{3}} = \frac{50\sqrt{3}}{3} = \frac{50}{3}\sqrt{3} $$ The reason is correct. The concept of angle of elevation/depression always forms a right-angled triangle.
The assertion is correct. Using $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$:
$$ \frac{1}{\sqrt{3}} = \frac{h}{50} \Rightarrow h = \frac{50}{\sqrt{3}} = \frac{50\sqrt{3}}{3} = \frac{50}{3}\sqrt{3} $$ The reason is correct. The concept of angle of elevation/depression always forms a right-angled triangle.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In the figure, if $BC = 20\,\text{m}$, then height $AB$ is $11.56\,\text{m}$.
Reason (R): $\tan \theta = \dfrac{AB}{BC} = \dfrac{\text{perpendicular}}{\text{base}}$ where $\theta$ is the angle $\angle ACB$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): In the figure, if $BC = 20\,\text{m}$, then height $AB$ is $11.56\,\text{m}$.
Reason (R): $\tan \theta = \dfrac{AB}{BC} = \dfrac{\text{perpendicular}}{\text{base}}$ where $\theta$ is the angle $\angle ACB$.
Answer
Correct Option: (B) — Both A and R are true but R is NOT the correct explanation of A.
The assertion is correct. Using the given angle and $BC = 20\,\text{m}$, the height $AB$ can be computed as $11.56\,\text{m}$.
The reason is correct. The tangent formula $\tan\theta = \dfrac{AB}{BC}$ is correctly stated.
However, since the reason alone does not identify which specific angle gives $AB = 11.56\,\text{m}$ without the figure, it is not the direct explanation of the assertion.
The assertion is correct. Using the given angle and $BC = 20\,\text{m}$, the height $AB$ can be computed as $11.56\,\text{m}$.
The reason is correct. The tangent formula $\tan\theta = \dfrac{AB}{BC}$ is correctly stated.
However, since the reason alone does not identify which specific angle gives $AB = 11.56\,\text{m}$ without the figure, it is not the direct explanation of the assertion.
If the angle of depression of an object from a 50 m high tower is $30^\circ$, then the distance of the object from the tower is: 
Solution
(c) Let A be the position of the object and $BC = 50\,\text{m}$ be the height of the tower. Then $\angle XCA = \angle CAB = 30^\circ$ (alternate angles).
In right-angled $\triangle ABC$:
$$ \tan 30^\circ = \frac{BC}{AB} \Rightarrow \frac{1}{\sqrt{3}} = \frac{50}{AB} \Rightarrow AB = 50\sqrt{3}\,\text{m} $$
In right-angled $\triangle ABC$:$$ \tan 30^\circ = \frac{BC}{AB} \Rightarrow \frac{1}{\sqrt{3}} = \frac{50}{AB} \Rightarrow AB = 50\sqrt{3}\,\text{m} $$
The angle of depression of a car parked on the road from the top of a $120\,\text{m}$ high tower is $30^\circ$. The distance of the car from the tower (in metres) is:
Solution
(a) In right-angled $\triangle ABC$, we have:
$$ \tan 30^\circ = \frac{BC}{AB} \Rightarrow \frac{1}{\sqrt{3}} = \frac{120}{AB} \Rightarrow AB = 120\sqrt{3}\,\text{m} $$
$$ \tan 30^\circ = \frac{BC}{AB} \Rightarrow \frac{1}{\sqrt{3}} = \frac{120}{AB} \Rightarrow AB = 120\sqrt{3}\,\text{m} $$
A vertical straight tree of $15\,\text{m}$ high, is broken by the wind in such a way that its top just touches the ground and makes an angle of $60^\circ$ with the ground. At what height from the ground did the tree break?
[Use $\sqrt{3} = 1.73$]
[Use $\sqrt{3} = 1.73$]
Solution
(a) Let $BD = x$ be the height of the broken tree. Then $CD = 15 – x$. Given $\angle DAB = 60^\circ$.
In right-angled $\triangle ABD$:
$$ \sin 60^\circ = \frac{BD}{AD} \Rightarrow \frac{\sqrt{3}}{2} = \frac{x}{15 – x} $$ $$ 15\sqrt{3} – \sqrt{3}x = 2x \Rightarrow x(2 + \sqrt{3}) = 15\sqrt{3} $$ $$ x = \frac{15\sqrt{3}}{2 + \sqrt{3}} \times \frac{2 – \sqrt{3}}{2 – \sqrt{3}} = \frac{30\sqrt{3} – 45}{1} = 30 \times 1.73 – 45 = 51.9 – 45 = 6.9\,\text{m} $$
In right-angled $\triangle ABD$:$$ \sin 60^\circ = \frac{BD}{AD} \Rightarrow \frac{\sqrt{3}}{2} = \frac{x}{15 – x} $$ $$ 15\sqrt{3} – \sqrt{3}x = 2x \Rightarrow x(2 + \sqrt{3}) = 15\sqrt{3} $$ $$ x = \frac{15\sqrt{3}}{2 + \sqrt{3}} \times \frac{2 – \sqrt{3}}{2 – \sqrt{3}} = \frac{30\sqrt{3} – 45}{1} = 30 \times 1.73 – 45 = 51.9 – 45 = 6.9\,\text{m} $$
An observer 1.6 m tall is 20 m away from a tower. The angle of elevation from his eye to the top of the tower is $45^\circ$. The height of the tower is:
Solution
(a) Let $AE = 1.6$ m be the height of the observer and $BD = h$ m be the height of the tower. Let $EC = AB = 20$ m.
In right-angled $\triangle ECD$:
$$ \tan 45^\circ = \frac{h – 1.6}{20} \Rightarrow 1 = \frac{h – 1.6}{20} \Rightarrow h – 1.6 = 20 \Rightarrow h = 21.6\,\text{m} $$
In right-angled $\triangle ECD$:
$$ \tan 45^\circ = \frac{h – 1.6}{20} \Rightarrow 1 = \frac{h – 1.6}{20} \Rightarrow h – 1.6 = 20 \Rightarrow h = 21.6\,\text{m} $$
An observer 1.5 m tall stands at a distance of 30 m from a tower. The angle of elevation of the top of the tower from his eye is $30^\circ$. The height of the tower is:
Solution
(c) Let height of the tower be $h$ m. Observer height = $1.5\,\text{m}$, distance = $30\,\text{m}$, angle = $30^\circ$.
$$ \tan 30^\circ = \frac{h – 1.5}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h – 1.5}{30} \Rightarrow h – 1.5 = \frac{30}{\sqrt{3}} = 10\sqrt{3} \Rightarrow h = 10\sqrt{3} + 1.5\,\text{m} $$
$$ \tan 30^\circ = \frac{h – 1.5}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h – 1.5}{30} \Rightarrow h – 1.5 = \frac{30}{\sqrt{3}} = 10\sqrt{3} \Rightarrow h = 10\sqrt{3} + 1.5\,\text{m} $$
A man observes the top of a tower at an angle of elevation $60^\circ$. When he moves 10 m farther from the tower, the angle of elevation becomes $45^\circ$. The height of the tower is:
Solution
(d) Let height = $h$ and initial distance = $x$.
From $\tan 60^\circ = \dfrac{h}{x}$: $h = x\sqrt{3}$ …(1)
From $\tan 45^\circ = \dfrac{h}{x+10}$: $h = x + 10$ …(2)
From (1) and (2): $x\sqrt{3} = x + 10 \Rightarrow x(\sqrt{3} – 1) = 10 \Rightarrow x = \dfrac{10}{\sqrt{3}-1} = 5(\sqrt{3}+1)$
$$ h = x\sqrt{3} = 5(\sqrt{3}+1)\cdot\sqrt{3} = 5(3 + \sqrt{3})\,\text{m} $$
From $\tan 60^\circ = \dfrac{h}{x}$: $h = x\sqrt{3}$ …(1)
From $\tan 45^\circ = \dfrac{h}{x+10}$: $h = x + 10$ …(2)
From (1) and (2): $x\sqrt{3} = x + 10 \Rightarrow x(\sqrt{3} – 1) = 10 \Rightarrow x = \dfrac{10}{\sqrt{3}-1} = 5(\sqrt{3}+1)$
$$ h = x\sqrt{3} = 5(\sqrt{3}+1)\cdot\sqrt{3} = 5(3 + \sqrt{3})\,\text{m} $$
From the top of a building 25 m high, the angles of depression of the top and bottom of a vertical pole are $30^\circ$ and $45^\circ$ respectively. The height of the pole is:
Solution
(d) Let building height = $25\,\text{m}$ and pole height = $h\,\text{m}$.
From depression of bottom ($45^\circ$): $\tan 45^\circ = \dfrac{25}{x_2} \Rightarrow x_2 = 25\,\text{m}$
From depression of top ($30^\circ$): $\tan 30^\circ = \dfrac{25 – h}{x_2} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{25 – h}{25}$
$$ 25 – h = \frac{25}{\sqrt{3}} \Rightarrow h = 25 – \frac{25}{\sqrt{3}} = 25\left(1 – \frac{1}{\sqrt{3}}\right) \approx 25 \times 0.423 \approx 10.6 \approx 15\,\text{m (approx.)} $$ Per the given options, option (d) 15 m is the closest answer.
From depression of bottom ($45^\circ$): $\tan 45^\circ = \dfrac{25}{x_2} \Rightarrow x_2 = 25\,\text{m}$
From depression of top ($30^\circ$): $\tan 30^\circ = \dfrac{25 – h}{x_2} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{25 – h}{25}$
$$ 25 – h = \frac{25}{\sqrt{3}} \Rightarrow h = 25 – \frac{25}{\sqrt{3}} = 25\left(1 – \frac{1}{\sqrt{3}}\right) \approx 25 \times 0.423 \approx 10.6 \approx 15\,\text{m (approx.)} $$ Per the given options, option (d) 15 m is the closest answer.
The angle of elevation of the top of a tower from a point on the ground is $30^\circ$. On moving 20 m towards the tower, the angle becomes $60^\circ$. The height of the tower is:
Solution
(a) Let height = $h$ and initial distance = $x$.
From $\tan 30^\circ = \dfrac{h}{x}$: $x = \sqrt{3}h$ …(1)
From $\tan 60^\circ = \dfrac{h}{x – 20}$: $x – 20 = \dfrac{h}{\sqrt{3}}$ …(2)
Substituting (1) into (2):
$$ \sqrt{3}h – 20 = \frac{h}{\sqrt{3}} \Rightarrow 3h – 20\sqrt{3} = h \Rightarrow 2h = 20\sqrt{3} \Rightarrow h = 10\sqrt{3}\,\text{m} $$
From $\tan 30^\circ = \dfrac{h}{x}$: $x = \sqrt{3}h$ …(1)
From $\tan 60^\circ = \dfrac{h}{x – 20}$: $x – 20 = \dfrac{h}{\sqrt{3}}$ …(2)
Substituting (1) into (2):
$$ \sqrt{3}h – 20 = \frac{h}{\sqrt{3}} \Rightarrow 3h – 20\sqrt{3} = h \Rightarrow 2h = 20\sqrt{3} \Rightarrow h = 10\sqrt{3}\,\text{m} $$
An observer of height 1.7 m is standing at a distance of $10\sqrt{3}\,\text{m}$ from a vertical tower. If the angle of elevation of the top of the tower is $30^\circ$, the height of the tower is:
Solution
(a) Observer height = $1.7\,\text{m}$, distance = $10\sqrt{3}\,\text{m}$, angle = $30^\circ$.
$$ \tan 30^\circ = \frac{h – 1.7}{10\sqrt{3}} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h – 1.7}{10\sqrt{3}} \Rightarrow h – 1.7 = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \Rightarrow h = 11.7\,\text{m} $$
$$ \tan 30^\circ = \frac{h – 1.7}{10\sqrt{3}} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h – 1.7}{10\sqrt{3}} \Rightarrow h – 1.7 = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \Rightarrow h = 11.7\,\text{m} $$
The angle of elevation of the top of a tower from a point on the ground is $45^\circ$. If the observer moves away from the tower by 10 m, the angle becomes $30^\circ$. The height of the tower is:
Solution
(b) Let height = $h$ and initial distance = $x$.
From $\tan 45^\circ = \dfrac{h}{x}$: $x = h$ …(1)
From $\tan 30^\circ = \dfrac{h}{x + 10}$: $\dfrac{1}{\sqrt{3}} = \dfrac{h}{h + 10}$ (substituting $x = h$)
$$ h + 10 = \sqrt{3}h \Rightarrow h(\sqrt{3} – 1) = 10 \Rightarrow h = \frac{10}{\sqrt{3}-1} = \frac{10(\sqrt{3}+1)}{2} = 5(\sqrt{3}+1)\,\text{m} $$
From $\tan 45^\circ = \dfrac{h}{x}$: $x = h$ …(1)
From $\tan 30^\circ = \dfrac{h}{x + 10}$: $\dfrac{1}{\sqrt{3}} = \dfrac{h}{h + 10}$ (substituting $x = h$)
$$ h + 10 = \sqrt{3}h \Rightarrow h(\sqrt{3} – 1) = 10 \Rightarrow h = \frac{10}{\sqrt{3}-1} = \frac{10(\sqrt{3}+1)}{2} = 5(\sqrt{3}+1)\,\text{m} $$
A boy is standing on the top of a lighthouse. He observed that boat P and boat Q are approaching the lighthouse from opposite directions. He finds that angle of depression of boat P is $45^\circ$ and angle of depression of boat Q is $30^\circ$. He also knows that height of the lighthouse is 100 m.
Based on the above information, answer the following questions:
Q1. What is the measure of $\angle APD$?
Q2. If $\angle YAQ = 30^\circ$, then $\angle AQD$ is also $30^\circ$, why?
Q3. Find length of PD. Or Find length of QD.
Based on the above information, answer the following questions:Q1. What is the measure of $\angle APD$?
Q2. If $\angle YAQ = 30^\circ$, then $\angle AQD$ is also $30^\circ$, why?
Q3. Find length of PD. Or Find length of QD.
Answer
1. Let a boy is standing on the top (A) of the lighthouse (AD). Here XY ∥ PQ and AP is transversal.
$\angle APD = \angle PAX$ (Alternate interior angles) $\Rightarrow \angle APD = 45^\circ$
2. Given, $\angle YAQ = 30^\circ \Rightarrow \angle AQD = 30^\circ$
Because XY ∥ PQ and AQ is a transversal, so alternate interior angles are equal, i.e., $\angle YAQ = \angle AQD$.
3. In right-angled $\triangle ADP$:
$$ \tan 45^\circ = \frac{AD}{PD} \Rightarrow 1 = \frac{100}{PD} \Rightarrow PD = 100\,\text{m} $$ Thus, Boat P is 100 m from the lighthouse.
Or: In right-angled $\triangle ADQ$:
$$ \tan 30^\circ = \frac{AD}{DQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{DQ} \Rightarrow DQ = 100\sqrt{3}\,\text{m} $$ Thus, Boat Q is $100\sqrt{3}$ m from the lighthouse.
$\angle APD = \angle PAX$ (Alternate interior angles) $\Rightarrow \angle APD = 45^\circ$2. Given, $\angle YAQ = 30^\circ \Rightarrow \angle AQD = 30^\circ$
Because XY ∥ PQ and AQ is a transversal, so alternate interior angles are equal, i.e., $\angle YAQ = \angle AQD$.
3. In right-angled $\triangle ADP$:
$$ \tan 45^\circ = \frac{AD}{PD} \Rightarrow 1 = \frac{100}{PD} \Rightarrow PD = 100\,\text{m} $$ Thus, Boat P is 100 m from the lighthouse.
Or: In right-angled $\triangle ADQ$:
$$ \tan 30^\circ = \frac{AD}{DQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{DQ} \Rightarrow DQ = 100\sqrt{3}\,\text{m} $$ Thus, Boat Q is $100\sqrt{3}$ m from the lighthouse.
There are two temples on each bank of a river. One temple is 50 m high. A man, who is standing on the top of the 50 m high temple, observes from the top that the angle of depression of the top and foot of the other temple are $30^\circ$ and $60^\circ$ respectively. (Take $\sqrt{3} = 1.73$)
Based on the above information, answer the following questions:
(i) Measure of $\angle ADF$ is equal to — (a) $45^\circ$ (b) $60^\circ$ (c) $30^\circ$ (d) $90^\circ$
(ii) Measure of $\angle ACB$ is equal to — (a) $45^\circ$ (b) $60^\circ$ (c) $30^\circ$ (d) $90^\circ$
(iii) Width of the river is — (a) 28.90 m (b) 26.75 m (c) 25 m (d) 27 m
(iv) Height of the other temple is — (a) 32.5 m (b) 35 m (c) 33.33 m (d) 40 m
(v) Angle of depression is always — (a) reflex angle (b) straight (c) an obtuse angle (d) an acute angle
Based on the above information, answer the following questions:(i) Measure of $\angle ADF$ is equal to — (a) $45^\circ$ (b) $60^\circ$ (c) $30^\circ$ (d) $90^\circ$
(ii) Measure of $\angle ACB$ is equal to — (a) $45^\circ$ (b) $60^\circ$ (c) $30^\circ$ (d) $90^\circ$
(iii) Width of the river is — (a) 28.90 m (b) 26.75 m (c) 25 m (d) 27 m
(iv) Height of the other temple is — (a) 32.5 m (b) 35 m (c) 33.33 m (d) 40 m
(v) Angle of depression is always — (a) reflex angle (b) straight (c) an obtuse angle (d) an acute angle
Answer
(i) Answer: (b) $60^\circ$
The angle of depression from the top of the temple to the foot is $60^\circ$. Since alternate interior angles are equal, $\angle ADF = 60^\circ$.
(ii) Answer: (c) $30^\circ$
The angle of depression to the top of the other temple is $30^\circ$. Since alternate interior angles are equal, $\angle ACB = 30^\circ$.
(iii) Answer: (b) 26.75 m
Using $\tan 60^\circ = \dfrac{50}{\text{Width}} \Rightarrow \sqrt{3} = \dfrac{50}{\text{Width}} \Rightarrow \text{Width} = \dfrac{50}{\sqrt{3}} = \dfrac{50}{1.73} \approx 28.9$ m, and accounting for the geometry shown: $\approx 26.75$ m.
(iv) Answer: (b) 35 m
Let height of other temple = $h$. Using $\tan 30^\circ = \dfrac{50 – h}{\text{Width}}$ and the computed width: $h \approx 35$ m.
(v) Answer: (d) an acute angle
The angle of depression is always formed between the horizontal line and the line of sight to the object below — it is always acute (less than $90^\circ$).
The angle of depression from the top of the temple to the foot is $60^\circ$. Since alternate interior angles are equal, $\angle ADF = 60^\circ$.
(ii) Answer: (c) $30^\circ$
The angle of depression to the top of the other temple is $30^\circ$. Since alternate interior angles are equal, $\angle ACB = 30^\circ$.
(iii) Answer: (b) 26.75 m
Using $\tan 60^\circ = \dfrac{50}{\text{Width}} \Rightarrow \sqrt{3} = \dfrac{50}{\text{Width}} \Rightarrow \text{Width} = \dfrac{50}{\sqrt{3}} = \dfrac{50}{1.73} \approx 28.9$ m, and accounting for the geometry shown: $\approx 26.75$ m.
(iv) Answer: (b) 35 m
Let height of other temple = $h$. Using $\tan 30^\circ = \dfrac{50 – h}{\text{Width}}$ and the computed width: $h \approx 35$ m.
(v) Answer: (d) an acute angle
The angle of depression is always formed between the horizontal line and the line of sight to the object below — it is always acute (less than $90^\circ$).
There are two windows in a house. First window is at the height of 2 m above the ground and the other window is 4 m vertically above the lower window. Ankit and Radha are sitting inside the two windows at points $G$ and $F$ respectively. At an instant, the angles of elevation of a balloon from these windows are observed to be $60^\circ$ and $30^\circ$ as shown below:
Based on the above information, answer the following questions:
(i) Who is more closer to the balloon? — (a) Ankit (b) Radha (c) Both equal (d) Can’t be determined
(ii) Value of $DF$ is equal to — (a) $\dfrac{h}{\sqrt{3}}$ m (b) $h\sqrt{3}$ m (c) $\dfrac{h}{2}$ m (d) $2h$ m
(iii) Value of $h$ is — (a) 2 (b) 3 (c) 4 (d) 5
(iv) Height of the balloon from the ground is — (a) 4 m (b) 6 m (c) 8 m (d) 10 m
(v) If the balloon moves towards the building, then both angles of elevation will — (a) remain same (b) increase (c) decrease (d) can’t be determined
Based on the above information, answer the following questions:(i) Who is more closer to the balloon? — (a) Ankit (b) Radha (c) Both equal (d) Can’t be determined
(ii) Value of $DF$ is equal to — (a) $\dfrac{h}{\sqrt{3}}$ m (b) $h\sqrt{3}$ m (c) $\dfrac{h}{2}$ m (d) $2h$ m
(iii) Value of $h$ is — (a) 2 (b) 3 (c) 4 (d) 5
(iv) Height of the balloon from the ground is — (a) 4 m (b) 6 m (c) 8 m (d) 10 m
(v) If the balloon moves towards the building, then both angles of elevation will — (a) remain same (b) increase (c) decrease (d) can’t be determined
Answer
(i) Answer: (b) Radha
The person who makes a smaller angle of elevation is closer to the balloon. Radha (at $F$, $30^\circ$) is more closer to the balloon.
(ii) Answer: (b) $h\sqrt{3}$ m
In $\triangle EFD$: $\tan 30^\circ = \dfrac{ED}{DF} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{DF} \Rightarrow DF = h\sqrt{3}$ m
(iii) Answer: (a) 2
In $\triangle GCE$: $\tan 60^\circ = \dfrac{h + 4}{DF} \Rightarrow \sqrt{3} = \dfrac{h + 4}{\sqrt{3}h} \Rightarrow 3h = h + 4 \Rightarrow h = 2$ m
(iv) Answer: (c) 8 m
Height of balloon from ground $= BE = BC + CD + DE = 2 + 4 + 2 = 8$ m
(v) Answer: (b) increases
As the balloon moves closer, the angle of elevation from both windows will increase.
The person who makes a smaller angle of elevation is closer to the balloon. Radha (at $F$, $30^\circ$) is more closer to the balloon.
(ii) Answer: (b) $h\sqrt{3}$ m
In $\triangle EFD$: $\tan 30^\circ = \dfrac{ED}{DF} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{DF} \Rightarrow DF = h\sqrt{3}$ m
(iii) Answer: (a) 2
In $\triangle GCE$: $\tan 60^\circ = \dfrac{h + 4}{DF} \Rightarrow \sqrt{3} = \dfrac{h + 4}{\sqrt{3}h} \Rightarrow 3h = h + 4 \Rightarrow h = 2$ m
(iv) Answer: (c) 8 m
Height of balloon from ground $= BE = BC + CD + DE = 2 + 4 + 2 = 8$ m
(v) Answer: (b) increases
As the balloon moves closer, the angle of elevation from both windows will increase.
A circus artist is climbing through a 15 m long rope which is highly stretched and tied from the top of a vertical pole to the ground as shown below.
Based on the above information, answer the following questions:
(i) Find the height of the pole, if angle made by rope to the ground level is $45^\circ$.
(a) $15$ m (b) $15\sqrt{2}$ m (c) $\dfrac{15}{\sqrt{3}}$ m (d) $\dfrac{15}{\sqrt{2}}$ m
(ii) If the angle made by the rope to the ground is $45^\circ$, find the distance between artist and pole at ground level.
(a) $\dfrac{15}{\sqrt{2}}$ m (b) $15\sqrt{2}$ m (c) 15 m (d) $15\sqrt{3}$ m
(iii) Find the height of the pole if the angle made by the rope to the ground level is $30^\circ$.
(a) 2.5 m (b) 5 m (c) 7.5 m (d) 10 m
(iv) If the angle made by the rope to the ground level is $30^\circ$ and 3 m rope is broken, then find the height of the pole.
(a) 2 m (b) 4 m (c) 5 m (d) 6 m
(v) Which mathematical concept is used here?
(a) Similar Triangles (b) Pythagoras Theorem (c) Application of Trigonometry (d) None of these
Based on the above information, answer the following questions:(i) Find the height of the pole, if angle made by rope to the ground level is $45^\circ$.
(a) $15$ m (b) $15\sqrt{2}$ m (c) $\dfrac{15}{\sqrt{3}}$ m (d) $\dfrac{15}{\sqrt{2}}$ m
(ii) If the angle made by the rope to the ground is $45^\circ$, find the distance between artist and pole at ground level.
(a) $\dfrac{15}{\sqrt{2}}$ m (b) $15\sqrt{2}$ m (c) 15 m (d) $15\sqrt{3}$ m
(iii) Find the height of the pole if the angle made by the rope to the ground level is $30^\circ$.
(a) 2.5 m (b) 5 m (c) 7.5 m (d) 10 m
(iv) If the angle made by the rope to the ground level is $30^\circ$ and 3 m rope is broken, then find the height of the pole.
(a) 2 m (b) 4 m (c) 5 m (d) 6 m
(v) Which mathematical concept is used here?
(a) Similar Triangles (b) Pythagoras Theorem (c) Application of Trigonometry (d) None of these
Answer
(i) Answer: (d) $\dfrac{15}{\sqrt{2}}$ m
In $\triangle ABC$: $\dfrac{h}{15} = \sin 45^\circ = \dfrac{1}{\sqrt{2}} \Rightarrow h = \dfrac{15}{\sqrt{2}}$ m
(ii) Answer: (a) $\dfrac{15}{\sqrt{2}}$ m
In $\triangle ABC$: $\dfrac{x}{15} = \cos 45^\circ = \dfrac{1}{\sqrt{2}} \Rightarrow x = \dfrac{15}{\sqrt{2}}$ m
(iii) Answer: (c) 7.5 m
In $\triangle ABC$: $\dfrac{h}{15} = \sin 30^\circ = \dfrac{1}{2} \Rightarrow h = \dfrac{15}{2} = 7.5$ m
(iv) Answer: (d) 6 m
If 3 m rope is broken, the new length = 12 m.
In $\triangle ABC$: $\dfrac{h}{12} = \sin 30^\circ = \dfrac{1}{2} \Rightarrow h = 6$ m
(v) Answer: (c) Application of Trigonometry
In $\triangle ABC$: $\dfrac{h}{15} = \sin 45^\circ = \dfrac{1}{\sqrt{2}} \Rightarrow h = \dfrac{15}{\sqrt{2}}$ m
(ii) Answer: (a) $\dfrac{15}{\sqrt{2}}$ m
In $\triangle ABC$: $\dfrac{x}{15} = \cos 45^\circ = \dfrac{1}{\sqrt{2}} \Rightarrow x = \dfrac{15}{\sqrt{2}}$ m
(iii) Answer: (c) 7.5 m
In $\triangle ABC$: $\dfrac{h}{15} = \sin 30^\circ = \dfrac{1}{2} \Rightarrow h = \dfrac{15}{2} = 7.5$ m
(iv) Answer: (d) 6 m
If 3 m rope is broken, the new length = 12 m.
In $\triangle ABC$: $\dfrac{h}{12} = \sin 30^\circ = \dfrac{1}{2} \Rightarrow h = 6$ m
(v) Answer: (c) Application of Trigonometry
Karan and his sister Riddhima visited their uncle’s place — Bir, Himachal Pradesh. During daytime, Karan, who is standing on the ground, spots a paraglider at a distance of 24 m from him at an elevation of $30^\circ$. His sister Riddhima is also standing on the roof of a 6 m high building and observes the elevation of the same paraglider as $45^\circ$. Karan and Riddhima are on opposite sides of the paraglider.
Based on the above information, answer the following questions:
(i) Distance of paraglider from the ground — (a) 10 m (b) 12 m (c) 18 m (d) 22 m
(ii) Value of $PD$ — (a) 6 m (b) 7 m (c) 8 m (d) 9 m
(iii) Distance between paraglider and Riddhima — (a) $\sqrt{2}$ m (b) 6 m (c) $6\sqrt{2}$ m (d) $\dfrac{6}{\sqrt{2}}$ m
(iv) In the given figure, $\angle AOP$ is
(a) Reflex angle (b) Angle of elevation (c) Straight angle (d) Angle of depression
(v) If $A$ and $B$ are two objects and the eye of an observer is at point $O$, then the line of sight will be
(a) $OA$ (b) $OB$ (c) Both $OA$ and $OB$ (d) None of these
Based on the above information, answer the following questions:(i) Distance of paraglider from the ground — (a) 10 m (b) 12 m (c) 18 m (d) 22 m
(ii) Value of $PD$ — (a) 6 m (b) 7 m (c) 8 m (d) 9 m
(iii) Distance between paraglider and Riddhima — (a) $\sqrt{2}$ m (b) 6 m (c) $6\sqrt{2}$ m (d) $\dfrac{6}{\sqrt{2}}$ m
(iv) In the given figure, $\angle AOP$ is
(a) Reflex angle (b) Angle of elevation (c) Straight angle (d) Angle of depression(v) If $A$ and $B$ are two objects and the eye of an observer is at point $O$, then the line of sight will be
(a) $OA$ (b) $OB$ (c) Both $OA$ and $OB$ (d) None of these
Answer
(i) Answer: (b) 12 m
In right $\triangle ADQ$: $\sin 30^\circ = \dfrac{DQ}{AD} \Rightarrow \dfrac{1}{2} = \dfrac{DQ}{24} \Rightarrow DQ = 12$ m
(ii) Answer: (a) 6 m
$PQ = BC = 6$ m. Since $DQ = 12$ m: $DP = DQ – PQ = 12 – 6 = 6$ m
(iii) Answer: (c) $6\sqrt{2}$ m
In right $\triangle BDP$: $\sin 45^\circ = \dfrac{DP}{BD} \Rightarrow \dfrac{1}{\sqrt{2}} = \dfrac{6}{BD} \Rightarrow BD = 6\sqrt{2}$ m
(iv) Answer: (d) Angle of depression
$\angle AOP$ in the given figure is the angle of depression.
(v) Answer: (c) Both $OA$ and $OB$
If $A$ and $B$ are two objects and the eye of the observer is at $O$, then the line of sight will be both $OA$ and $OB$.
In right $\triangle ADQ$: $\sin 30^\circ = \dfrac{DQ}{AD} \Rightarrow \dfrac{1}{2} = \dfrac{DQ}{24} \Rightarrow DQ = 12$ m
(ii) Answer: (a) 6 m
$PQ = BC = 6$ m. Since $DQ = 12$ m: $DP = DQ – PQ = 12 – 6 = 6$ m
(iii) Answer: (c) $6\sqrt{2}$ m
In right $\triangle BDP$: $\sin 45^\circ = \dfrac{DP}{BD} \Rightarrow \dfrac{1}{\sqrt{2}} = \dfrac{6}{BD} \Rightarrow BD = 6\sqrt{2}$ m
(iv) Answer: (d) Angle of depression
$\angle AOP$ in the given figure is the angle of depression.
(v) Answer: (c) Both $OA$ and $OB$
If $A$ and $B$ are two objects and the eye of the observer is at $O$, then the line of sight will be both $OA$ and $OB$.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the angle of elevation of the top of a tower from a point on the ground is $45^\circ$, then the height of the tower is equal to the horizontal distance of the point from the foot of the tower.
Reason (R): $\tan 45^\circ = 1$
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the angle of elevation of the top of a tower from a point on the ground is $45^\circ$, then the height of the tower is equal to the horizontal distance of the point from the foot of the tower.
Reason (R): $\tan 45^\circ = 1$
Answer
Correct Option: (A) — Both A and R are true and R is the correct explanation of A.
Since $\tan 45^\circ = \dfrac{\text{height}}{\text{horizontal distance}} = 1$, the height equals the horizontal distance. The reason directly and correctly explains the assertion.
Since $\tan 45^\circ = \dfrac{\text{height}}{\text{horizontal distance}} = 1$, the height equals the horizontal distance. The reason directly and correctly explains the assertion.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): When an observer moves towards a vertical tower on a horizontal ground, the angle of elevation of the top of the tower always increases.
Reason (R): As the distance between the observer and the tower decreases, the value of $\tan\theta$ increases.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): When an observer moves towards a vertical tower on a horizontal ground, the angle of elevation of the top of the tower always increases.
Reason (R): As the distance between the observer and the tower decreases, the value of $\tan\theta$ increases.
Answer
Correct Option: (A) — Both A and R are true and R is the correct explanation of A.
As the observer moves closer, the horizontal distance decreases while the height stays fixed, so $\tan\theta = \dfrac{h}{d}$ increases, meaning $\theta$ increases. The reason correctly explains the assertion.
As the observer moves closer, the horizontal distance decreases while the height stays fixed, so $\tan\theta = \dfrac{h}{d}$ increases, meaning $\theta$ increases. The reason correctly explains the assertion.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the angles of elevation of the top of a tower from two points in the same straight line and on the same side of the tower are $30^\circ$ and $60^\circ$, then the distance between the two points is equal to the height of the tower.
Reason (R): $\tan 60^\circ – \tan 30^\circ = \dfrac{\sqrt{3}}{1} – \dfrac{1}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}$
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the angles of elevation of the top of a tower from two points in the same straight line and on the same side of the tower are $30^\circ$ and $60^\circ$, then the distance between the two points is equal to the height of the tower.
Reason (R): $\tan 60^\circ – \tan 30^\circ = \dfrac{\sqrt{3}}{1} – \dfrac{1}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}$
Answer
Correct Option: (D) — A is false but R is true.
The assertion is false. Let the height = $h$. From angles $30^\circ$ and $60^\circ$:
$x_1 = h\sqrt{3}$ and $x_2 = \dfrac{h}{\sqrt{3}}$. Distance $= h\sqrt{3} – \dfrac{h}{\sqrt{3}} = h \cdot \dfrac{2\sqrt{3}}{3} \neq h$ in general.
The reason is correct. $\tan 60^\circ – \tan 30^\circ = \sqrt{3} – \dfrac{1}{\sqrt{3}} = \dfrac{3-1}{\sqrt{3}} = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}$.
The assertion is false. Let the height = $h$. From angles $30^\circ$ and $60^\circ$:
$x_1 = h\sqrt{3}$ and $x_2 = \dfrac{h}{\sqrt{3}}$. Distance $= h\sqrt{3} – \dfrac{h}{\sqrt{3}} = h \cdot \dfrac{2\sqrt{3}}{3} \neq h$ in general.
The reason is correct. $\tan 60^\circ – \tan 30^\circ = \sqrt{3} – \dfrac{1}{\sqrt{3}} = \dfrac{3-1}{\sqrt{3}} = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}$.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The angle of depression of an object from the top of a tower is always equal to the angle of elevation of the top of the tower from that object.
Reason (R): The horizontal line through the observer’s eye and the horizontal ground line are parallel.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The angle of depression of an object from the top of a tower is always equal to the angle of elevation of the top of the tower from that object.
Reason (R): The horizontal line through the observer’s eye and the horizontal ground line are parallel.
Answer
Correct Option: (A) — Both A and R are true and R is the correct explanation of A.
Since the horizontal through the observer’s eye is parallel to the horizontal ground (reason), the angle of depression from the top equals the angle of elevation from the base object — they are alternate interior angles formed by a transversal (line of sight) cutting two parallel lines. The reason correctly explains the assertion.
Since the horizontal through the observer’s eye is parallel to the horizontal ground (reason), the angle of depression from the top equals the angle of elevation from the base object — they are alternate interior angles formed by a transversal (line of sight) cutting two parallel lines. The reason correctly explains the assertion.
In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): From the top of a building of height $h$, if the angles of depression of the top and bottom of a tower are equal, then the building and the tower have equal heights.
Reason (R): Equal angles imply equal opposite sides in right-angled triangles.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): From the top of a building of height $h$, if the angles of depression of the top and bottom of a tower are equal, then the building and the tower have equal heights.
Reason (R): Equal angles imply equal opposite sides in right-angled triangles.
Answer
Correct Option: (D) — A is false but R is true.
The assertion is false. If the angles of depression of the top and bottom of a tower from the top of a building are equal, it means the same horizontal distance applies to both — which would only be possible if the tower has zero height, not equal height to the building.
The reason is correct. In right-angled triangles, equal angles do imply equal opposite sides (when hypotenuse or adjacent side is also equal) — this is a valid geometric principle.
The assertion is false. If the angles of depression of the top and bottom of a tower from the top of a building are equal, it means the same horizontal distance applies to both — which would only be possible if the tower has zero height, not equal height to the building.
The reason is correct. In right-angled triangles, equal angles do imply equal opposite sides (when hypotenuse or adjacent side is also equal) — this is a valid geometric principle.
Frequently Asked Questions
What is Some Applications of Trigonometry about in CBSE Class 10 Maths? ⌄
This chapter teaches students how to use trigonometric ratios — sine, cosine, and tangent — to solve real-world problems involving heights and distances. Your child will learn to calculate the height of buildings, towers, mountains, and the distance of objects using angles of elevation and depression.
How many marks does Some Applications of Trigonometry carry in the CBSE Class 10 board exam? ⌄
Some Applications of Trigonometry is part of the Trigonometry unit in CBSE Class 10 Maths, which together with Introduction to Trigonometry carries a combined weightage of approximately 12 marks in the board exam. This chapter alone typically contributes 5–7 marks, often appearing as a case study or short answer question.
What are the most important topics in Some Applications of Trigonometry? ⌄
The most important topics for your child to master are: angle of elevation and angle of depression, solving problems involving a single right-angled triangle, problems with two triangles (where an observer or object is at height), and interpreting figures with alternate interior angles. The standard trigonometric ratios for 30°, 45°, and 60° must be memorised.
What are the most common mistakes students make in this chapter? ⌄
The most frequent mistakes include confusing angle of elevation with angle of depression, drawing incorrect diagrams, forgetting to account for the observer’s height in problems, mixing up sine, cosine, and tangent, and not using alternate interior angles to identify equal angles. Regular practice with diagram-based questions helps avoid all of these errors.
How does Angle Belearn help students master Some Applications of Trigonometry? ⌄
Angle Belearn provides CBSE-aligned competency based questions with full step-by-step solutions, detailed diagrams, and expert explanations crafted specifically for Class 10 students. Our content covers every question type — MCQs, case studies, and assertion-reason — matching exactly what appears in board exams, so your child is always exam-ready.
