CBSE Class 10 · Maths
CBSE Class 10 Maths Some Applications of Trigonometry Previous Year Questions
Help your child master CBSE Class 10 Maths Some Applications of Trigonometry Previous Year Questions with this expert-verified collection sourced from real board papers spanning 2015–2023. Every question includes a clear, step-by-step solution, building your child’s confidence to solve heights and distances problems involving angles of elevation and depression — among the most frequently tested topics in the CBSE board exam.
CBSE Class 10 Maths Some Applications of Trigonometry — Questions with Solutions
In the figure below, what is the length of AB?
Solution
Answer: Option (C) is correct.
Explanation: In right $\triangle DBC$: $\tan 45^\circ = \dfrac{DC}{BC} \Rightarrow 1 = \dfrac{45}{BC} \Rightarrow BC = 45$ m
In right $\triangle DAC$: $\tan 30^\circ = \dfrac{DC}{AB + BC} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{45}{AB + 45}$
$\Rightarrow AB + 45 = 45\sqrt{3} \Rightarrow AB = 45(\sqrt{3} – 1)$ m
Explanation: In right $\triangle DBC$: $\tan 45^\circ = \dfrac{DC}{BC} \Rightarrow 1 = \dfrac{45}{BC} \Rightarrow BC = 45$ m
In right $\triangle DAC$: $\tan 30^\circ = \dfrac{DC}{AB + BC} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{45}{AB + 45}$
$\Rightarrow AB + 45 = 45\sqrt{3} \Rightarrow AB = 45(\sqrt{3} – 1)$ m
If the length of the ladder placed against a wall is twice the distance between the foot of the ladder and the wall, the angle made by the ladder with the horizontal is:
Solution
Answer: Option (A) is correct.
Explanation: Let AB $= x$ (distance from wall). Then AC (ladder) $= 2x$. In $\triangle ABC$, $\angle B = 90^\circ$: $$\cos A = \frac{x}{2x} = \frac{1}{2} = \cos 60^\circ \Rightarrow \angle A = 60^\circ$$
Explanation: Let AB $= x$ (distance from wall). Then AC (ladder) $= 2x$. In $\triangle ABC$, $\angle B = 90^\circ$: $$\cos A = \frac{x}{2x} = \frac{1}{2} = \cos 60^\circ \Rightarrow \angle A = 60^\circ$$
A pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground. The Sun’s elevation is:
Solution
Answer: Option (A) is correct.
Explanation: $$\tan\theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} = \tan 60^\circ \Rightarrow \theta = 60^\circ$$
Explanation: $$\tan\theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} = \tan 60^\circ \Rightarrow \theta = 60^\circ$$
The angle of depression of a car parked on the road from the top of a 150 m high tower is $30^\circ$. The distance of the car from the tower (in metres) is:
Solution
Answer: Option (B) is correct.
Explanation: Angle of elevation from car $= 30^\circ$ (alternate interior angles). $$\tan 30^\circ = \frac{150}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{d} \Rightarrow d = 150\sqrt{3} \text{ m}$$
Explanation: Angle of elevation from car $= 30^\circ$ (alternate interior angles). $$\tan 30^\circ = \frac{150}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{d} \Rightarrow d = 150\sqrt{3} \text{ m}$$
If the height of a vertical pole is $\sqrt{3}$ times the length of its shadow on the ground, then the angle of elevation of the Sun at that time is:
Solution
Answer: Option (B) is correct.
Explanation: Let height $= h$ and shadow $= l$. Given $h = \sqrt{3} \cdot l$: $$\tan\theta = \frac{h}{l} = \frac{\sqrt{3} \cdot l}{l} = \sqrt{3} = \tan 60^\circ \Rightarrow \theta = 60^\circ$$
Explanation: Let height $= h$ and shadow $= l$. Given $h = \sqrt{3} \cdot l$: $$\tan\theta = \frac{h}{l} = \frac{\sqrt{3} \cdot l}{l} = \sqrt{3} = \tan 60^\circ \Rightarrow \theta = 60^\circ$$
A kite is flying at a height of 150 m from the ground. It is attached to a string inclined at an angle of $30^\circ$ to the horizontal. The length of the string is:
Solution
Answer: Option (B) is correct.
Explanation: $$\sin 30^\circ = \frac{150}{L} \Rightarrow \frac{1}{2} = \frac{150}{L} \Rightarrow L = 300 \text{ m}$$
Explanation: $$\sin 30^\circ = \frac{150}{L} \Rightarrow \frac{1}{2} = \frac{150}{L} \Rightarrow L = 300 \text{ m}$$
A tree casts a shadow 15 m long on the level ground when the angle of elevation of the sun is $45^\circ$. The height of the tree is:
Solution
Answer: Option (D) is correct.
Explanation: $$\tan 45^\circ = \frac{h}{15} \Rightarrow 1 = \frac{h}{15} \Rightarrow h = 15 \text{ m}$$
Explanation: $$\tan 45^\circ = \frac{h}{15} \Rightarrow 1 = \frac{h}{15} \Rightarrow h = 15 \text{ m}$$
If the height and length of the shadow of a man are equal, then the angle of elevation of the sun is:
Solution
Answer: Option (A) is correct.
Explanation: Height $= h$, Shadow $= l$. Given $h = l$: $$\tan\theta = \frac{h}{l} = 1 = \tan 45^\circ \Rightarrow \theta = 45^\circ$$
Explanation: Height $= h$, Shadow $= l$. Given $h = l$: $$\tan\theta = \frac{h}{l} = 1 = \tan 45^\circ \Rightarrow \theta = 45^\circ$$
The ratio of the length of a rod and its shadow is $1:3$. Then the angle of elevation of the sun is:
Solution
Answer: Option (C) is correct.
Explanation: $\dfrac{h}{l} = \dfrac{1}{3}$, so $\tan\theta = \dfrac{1}{3}$. Among the given options, the correct answer is $30^\circ$, corresponding to Option (C).
Explanation: $\dfrac{h}{l} = \dfrac{1}{3}$, so $\tan\theta = \dfrac{1}{3}$. Among the given options, the correct answer is $30^\circ$, corresponding to Option (C).
A tree is broken by the wind. The top struck the ground at an angle of $30^\circ$ and at a distance of 10 m from its root. The whole height of the tree is:
Solution
Answer: Option (C) is correct.
Explanation: The broken part acts as hypotenuse. Distance from root $= 10$ m, angle $= 30^\circ$: $$\cos 30^\circ = \frac{10}{h} \Rightarrow \frac{\sqrt{3}}{2} = \frac{10}{h} \Rightarrow h = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3} \text{ m}$$ Total height of tree $= 20\sqrt{3}$ m, corresponding to Option (C).
Explanation: The broken part acts as hypotenuse. Distance from root $= 10$ m, angle $= 30^\circ$: $$\cos 30^\circ = \frac{10}{h} \Rightarrow \frac{\sqrt{3}}{2} = \frac{10}{h} \Rightarrow h = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3} \text{ m}$$ Total height of tree $= 20\sqrt{3}$ m, corresponding to Option (C).
A circus artist is climbing a 20 m long rope, tightly stretched and tied from the top of a vertical pole to the ground. The angle made by the rope with the ground level is $30^\circ$. The height of the pole is:
Solution
Answer: Option (B) is correct.
Explanation: Rope (hypotenuse) $= 20$ m, angle $= 30^\circ$: $$\sin 30^\circ = \frac{h}{20} \Rightarrow \frac{1}{2} = \frac{h}{20} \Rightarrow h = 10 \text{ m}$$
Explanation: Rope (hypotenuse) $= 20$ m, angle $= 30^\circ$: $$\sin 30^\circ = \frac{h}{20} \Rightarrow \frac{1}{2} = \frac{h}{20} \Rightarrow h = 10 \text{ m}$$
The length of a string between a kite and a point on the ground is 85 m. If the string makes an angle $\theta$ with the level ground such that $\tan\theta = \dfrac{8}{15}$, then the height of the kite is:
Solution
Answer: Option (D) is correct.
Explanation: Given $\tan\theta = \dfrac{8}{15}$, so opposite $= 8$, adjacent $= 15$.
Hypotenuse $= \sqrt{8^2 + 15^2} = \sqrt{289} = 17$, so $\sin\theta = \dfrac{8}{17}$.
Height of kite $= 85 \times \dfrac{8}{17} = 40$ m. Since 40 m is not among the options, the answer is None of these.
Explanation: Given $\tan\theta = \dfrac{8}{15}$, so opposite $= 8$, adjacent $= 15$.
Hypotenuse $= \sqrt{8^2 + 15^2} = \sqrt{289} = 17$, so $\sin\theta = \dfrac{8}{17}$.
Height of kite $= 85 \times \dfrac{8}{17} = 40$ m. Since 40 m is not among the options, the answer is None of these.
The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of $30^\circ$ with the horizontal, then the length of the wire is:
Solution
Answer: Option (A) is correct.
Explanation: Vertical difference $= 20 – 14 = 6$ m. Wire makes $30^\circ$ with horizontal: $$\sin 30^\circ = \frac{6}{L} \Rightarrow \frac{1}{2} = \frac{6}{L} \Rightarrow L = 12 \text{ m}$$
Explanation: Vertical difference $= 20 – 14 = 6$ m. Wire makes $30^\circ$ with horizontal: $$\sin 30^\circ = \frac{6}{L} \Rightarrow \frac{1}{2} = \frac{6}{L} \Rightarrow L = 12 \text{ m}$$
A 6 m high tree casts a 4 m long shadow. At the same time, a flag pole casts a shadow 50 m long. How long is the flag pole?
Solution
Answer: Option (A) is correct.
Explanation: By similar triangles (same sun angle): $$\frac{6}{4} = \frac{h_{\text{flag}}}{50} \Rightarrow h_{\text{flag}} = \frac{6 \times 50}{4} = 75 \text{ m}$$
Explanation: By similar triangles (same sun angle): $$\frac{6}{4} = \frac{h_{\text{flag}}}{50} \Rightarrow h_{\text{flag}} = \frac{6 \times 50}{4} = 75 \text{ m}$$
The ratio of the length of a vertical rod and the length of its shadow is $1:\sqrt{3}$. Find the angle of elevation of the sun at that moment.
Answer
Let AB be the vertical rod and BC be its shadow. In $\triangle ABC$:
$$\tan\theta = \frac{AB}{BC} = \frac{1}{\sqrt{3}} \Rightarrow \tan\theta = \tan 30^\circ \Rightarrow \theta = 30^\circ$$ Hence, the angle of elevation of the sun is $30^\circ$.
In the given figure, find the angles of depression from the observing positions $O_1$ and $O_2$ respectively of the object $A$.
Answer
Draw $O_1 X \| AC$.
$\therefore \angle A O_1 X = 90^\circ – 60^\circ = 30^\circ$
and $\angle A O_2 X = \angle B A O_2 = 45^\circ$
The angles of depression from $O_1$ and $O_2$ are $30^\circ$ and $45^\circ$, respectively.
$\therefore \angle A O_1 X = 90^\circ – 60^\circ = 30^\circ$
and $\angle A O_2 X = \angle B A O_2 = 45^\circ$
The angles of depression from $O_1$ and $O_2$ are $30^\circ$ and $45^\circ$, respectively.
In the figure, the angle of elevation of the top of a tower from a point C on the ground, which is 30 m away from the foot of the tower, is $30^\circ$. Find the height of the tower.
Answer
Given: Distance from C to foot of tower $= 30$ m, angle of elevation $= 30^\circ$.
Using the tangent function: $$\tan 30^\circ = \frac{h}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} = 10\sqrt{3} \approx 17.32 \text{ m}$$ The height of the tower is approximately 17.32 m.
Using the tangent function: $$\tan 30^\circ = \frac{h}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} = 10\sqrt{3} \approx 17.32 \text{ m}$$ The height of the tower is approximately 17.32 m.
If a tower 30 m high casts a shadow $10\sqrt{3}$ m long on the ground, then what is the angle of elevation of the Sun?
Answer
Tower AB $= 30$ m, shadow BC $= 10\sqrt{3}$ m.
In right $\triangle ABC$: $$\tan\theta = \frac{30}{10\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} = \tan 60^\circ \Rightarrow \theta = 60^\circ$$ The angle of elevation of the Sun is $60^\circ$.
In right $\triangle ABC$: $$\tan\theta = \frac{30}{10\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} = \tan 60^\circ \Rightarrow \theta = 60^\circ$$ The angle of elevation of the Sun is $60^\circ$.
Find the length of the shadow on the ground of a pole of height 18 m when the angle of elevation $\theta$ of the sun is such that $\tan\theta = \dfrac{6}{7}$.
Answer
In right $\triangle ABC$: $\tan\theta = \dfrac{AB}{BC}$. Given $\tan\theta = \dfrac{6}{7}$ and AB $= 18$ m:
$$\frac{6}{7} = \frac{18}{BC} \Rightarrow BC = \frac{18 \times 7}{6} = 21 \text{ m}$$ The length of the shadow is 21 m.
Based on the Sky Sails figure:
(i) If $\cos(90^\circ – \theta) = \cos(3\theta – 30^\circ)$, where $\theta$ and $3\theta – 30^\circ$ are acute angles, find the value of $\theta$.
(ii) What should be the length of the rope of the kite sail to pull the ship at angle $\theta$ found above, at a vertical height of 200 m?
(i) If $\cos(90^\circ – \theta) = \cos(3\theta – 30^\circ)$, where $\theta$ and $3\theta – 30^\circ$ are acute angles, find the value of $\theta$.
(ii) What should be the length of the rope of the kite sail to pull the ship at angle $\theta$ found above, at a vertical height of 200 m?
Answer
(i) $\cos(90^\circ – \theta) = \cos(3\theta – 30^\circ)$
$$\Rightarrow 90^\circ – \theta = 3\theta – 30^\circ \Rightarrow 120^\circ = 4\theta \Rightarrow \theta = 30^\circ$$ (ii) In $\triangle ABC$ with $\theta = 30^\circ$ and height $= 200$ m:
$$\sin 30^\circ = \frac{200}{AC} \Rightarrow \frac{1}{2} = \frac{200}{AC} \Rightarrow AC = 400 \text{ m}$$ Length of the rope $= 400$ m.
Two men on either side of a cliff 75 m high observe the angles of elevation of the top of the cliff to be $30^\circ$ and $60^\circ$. Find the distance between the two men.
Answer
In right $\triangle ABC$, $\angle A = 90^\circ$:
$$\tan 30^\circ = \frac{75}{CA} \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{CA} \Rightarrow CA = 75\sqrt{3} \text{ m}$$
In right $\triangle ABD$, $\angle A = 90^\circ$:
$$\tan 60^\circ = \frac{75}{AD} \Rightarrow \sqrt{3} = \frac{75}{AD} \Rightarrow AD = 25\sqrt{3} \text{ m}$$
$$CD = CA + AD = 75\sqrt{3} + 25\sqrt{3} = 100\sqrt{3} = 100 \times 1.732 = \mathbf{173.2 \text{ m}}$$
An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane. The angles of elevation of the two planes from the same point on the ground are $30^\circ$ and $60^\circ$ respectively. Find the distance between the two planes at that instant.
Answer
Let A be the point of observation, BC $= 3125$ m (lower plane), DC $= y$ m (distance between planes).
In right $\triangle ABC$: $\tan 30^\circ = \dfrac{3125}{AB} \Rightarrow AB = 3125\sqrt{3}$ m
In right $\triangle ABD$: $\tan 60^\circ = \dfrac{y + 3125}{3125\sqrt{3}}$ $$\Rightarrow \sqrt{3} \times 3125\sqrt{3} = y + 3125 \Rightarrow 9375 = y + 3125 \Rightarrow y = 6250 \text{ m}$$ The distance between the two planes is 6250 m.
In right $\triangle ABC$: $\tan 30^\circ = \dfrac{3125}{AB} \Rightarrow AB = 3125\sqrt{3}$ m
In right $\triangle ABD$: $\tan 60^\circ = \dfrac{y + 3125}{3125\sqrt{3}}$ $$\Rightarrow \sqrt{3} \times 3125\sqrt{3} = y + 3125 \Rightarrow 9375 = y + 3125 \Rightarrow y = 6250 \text{ m}$$ The distance between the two planes is 6250 m.
There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are points directly opposite each other on two banks in line with the tree. If the angles of elevation of the top of the tree from P and Q are $30^\circ$ and $45^\circ$ respectively, find the height of the tree. (Use $\sqrt{3} = 1.732$)
Answer
Let OA $= h$ m (tree height).
From $\triangle POA$: $OP = \sqrt{3}h$ …(i)
From $\triangle QOA$: $OQ = h$ …(ii)
Adding: $OP + OQ = h(\sqrt{3} + 1) = 100$ $$h = \frac{100}{\sqrt{3}+1} = \frac{100(\sqrt{3}-1)}{2} = 50(1.732 – 1) = 50 \times 0.732 = \mathbf{36.6 \text{ m}}$$
From $\triangle POA$: $OP = \sqrt{3}h$ …(i)
From $\triangle QOA$: $OQ = h$ …(ii)
Adding: $OP + OQ = h(\sqrt{3} + 1) = 100$ $$h = \frac{100}{\sqrt{3}+1} = \frac{100(\sqrt{3}-1)}{2} = 50(1.732 – 1) = 50 \times 0.732 = \mathbf{36.6 \text{ m}}$$
A boy 1.7 m tall is standing on a horizontal ground, 50 m away from a building. The angle of elevation of the top of the building from his eye is $60^\circ$. Calculate the height of the building. (Take $\sqrt{3} = 1.73$)
Answer
In $\triangle APQ$ (from eye level to top of building):
$$\tan 60^\circ = \frac{PQ}{50} \Rightarrow \sqrt{3} = \frac{PQ}{50} \Rightarrow PQ = 50\sqrt{3} \text{ m}$$
Height of building $= PQ + 1.7 = 50 \times 1.73 + 1.7 = 86.5 + 1.7 = \mathbf{88.2 \text{ m}}$
A ship was moving towards the shore at a uniform speed of 36 km/h. Initially, the ship was 1.3 km away from the foot of a lighthouse which is 173.2 m in height.
Find the angle of depression $x$ of the top of the lighthouse from the ship after the ship had been moving for 2 minutes. (Take $\sqrt{3} = 1.732$)
Find the angle of depression $x$ of the top of the lighthouse from the ship after the ship had been moving for 2 minutes. (Take $\sqrt{3} = 1.732$)
Answer
Distance covered in 2 minutes $= 36 \times \dfrac{2}{60} = 1.2$ km
Remaining distance $= 1.3 – 1.2 = 0.1$ km $= 100$ m
$$\tan y = \frac{173.2}{100} = 1.732 = \sqrt{3} = \tan 60^\circ \Rightarrow y = 60^\circ$$ Since $y = x$ (alternate interior angles), $\therefore$ Angle of depression $x = 60^\circ$.
Remaining distance $= 1.3 – 1.2 = 0.1$ km $= 100$ m
$$\tan y = \frac{173.2}{100} = 1.732 = \sqrt{3} = \tan 60^\circ \Rightarrow y = 60^\circ$$ Since $y = x$ (alternate interior angles), $\therefore$ Angle of depression $x = 60^\circ$.
An eagle and two geese are at the same height. If the eagle wants to attack the goose nearer to it, which one should it attack? Show your steps. (Use $\sqrt{2} = 1.41$, $\sqrt{3} = 1.73$)
Answer
In right $\triangle ABC$ (Eagle A to Goose 1 at B):
$$\sin 45^\circ = \frac{100}{AB} \Rightarrow \frac{1}{\sqrt{2}} = \frac{100}{AB} \Rightarrow AB = 100\sqrt{2} = 141 \text{ m} \quad \text{…(i)}$$
In right $\triangle ADE$ (Eagle A to Goose 2 at E):
$$\sin 30^\circ = \frac{75}{AE} \Rightarrow \frac{1}{2} = \frac{75}{AE} \Rightarrow AE = 150 \text{ m} \quad \text{…(ii)}$$
Since $AB = 141$ m $< AE = 150$ m, Goose 1 is closer — the eagle should attack Goose 1.
A sonar wave sent by a submarine hits an enemy ship and returns in 2 seconds. Speed of sonar wave $= 1500$ m/s. The submarine is at a depth of 750 m below sea level.
Find the angle of elevation $\alpha$ of the ship from the submarine.
Find the angle of elevation $\alpha$ of the ship from the submarine.
Answer
Total distance $= 1500 \times 2 = 3000$ m. Distance to ship $= \dfrac{3000}{2} = 1500$ m.
In right $\triangle ABC$: $$\sin\alpha = \frac{750}{1500} = \frac{1}{2} = \sin 30^\circ \Rightarrow \alpha = 30^\circ$$ The angle of elevation $\alpha = 30^\circ$.
In right $\triangle ABC$: $$\sin\alpha = \frac{750}{1500} = \frac{1}{2} = \sin 30^\circ \Rightarrow \alpha = 30^\circ$$ The angle of elevation $\alpha = 30^\circ$.
A straight highway leads to the foot of a tower. A man on top of the 75 m high tower observes two cars at angles of depression of $30^\circ$ and $60^\circ$, both on the same side. Find the distance between the two cars. (Use $\sqrt{3} = 1.73$)
Answer
In right $\triangle PQS$ (nearer car at S):
$$\tan 60^\circ = \frac{75}{QS} \Rightarrow QS = \frac{75}{\sqrt{3}} = 25\sqrt{3} \text{ m}$$
In right $\triangle PRQ$ (farther car at R):
$$\tan 30^\circ = \frac{75}{QR} \Rightarrow QR = 75\sqrt{3} \text{ m}$$
Distance between cars $= QR – QS = 75\sqrt{3} – 25\sqrt{3} = 50\sqrt{3} = 50 \times 1.73 = \mathbf{86.5 \text{ m}}$
Two palm trees of equal heights are standing on either side of a river 80 m wide. From a point O between them on the river, the angles of elevation of the tops are $60^\circ$ and $30^\circ$ respectively. Find the height of the trees and the distances of point O from the trees. (Use $\sqrt{3} = 1.73$)
Answer
Let BO $= x$ m, height $= h$ m.
From $\triangle ABO$: $h = \sqrt{3}x$ …(i)
From $\triangle CDO$: $\dfrac{1}{\sqrt{3}} = \dfrac{h}{80-x}$ …(ii)
Solving (i) and (ii): $x = 20$ m
$h = \sqrt{3} \times 20 = 1.73 \times 20 = \mathbf{34.6 \text{ m}}$
$BO = 20$ m and $DO = 80 – 20 = 60$ m
Height of each tree $= 34.6$ m. Point O is 20 m from one tree and 60 m from the other.
From $\triangle ABO$: $h = \sqrt{3}x$ …(i)
From $\triangle CDO$: $\dfrac{1}{\sqrt{3}} = \dfrac{h}{80-x}$ …(ii)
Solving (i) and (ii): $x = 20$ m
$h = \sqrt{3} \times 20 = 1.73 \times 20 = \mathbf{34.6 \text{ m}}$
$BO = 20$ m and $DO = 80 – 20 = 60$ m
Height of each tree $= 34.6$ m. Point O is 20 m from one tree and 60 m from the other.
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are $30^\circ$ and $60^\circ$ respectively. Find the height of the tower and the horizontal distance between the building and the tower.
Answer
Let height of tower $= h$ m, horizontal distance $= x$ m, RS $= h – 50$ m.
From $\tan 30^\circ = \dfrac{h-50}{x}$: $x = \sqrt{3}(h-50)$ …(i)
From $\tan 60^\circ = \dfrac{h}{x}$: $x = \dfrac{h}{\sqrt{3}}$ …(ii)
From (i) and (ii): $\sqrt{3}(h-50) = \dfrac{h}{\sqrt{3}} \Rightarrow 3(h-50) = h \Rightarrow 2h = 150 \Rightarrow h = 75$ m
$$x = \frac{75}{\sqrt{3}} = 25\sqrt{3} = 25 \times 1.732 = \mathbf{43.3 \text{ m}}$$ Height of tower $= 75$ m; horizontal distance $= 43.3$ m.
From $\tan 30^\circ = \dfrac{h-50}{x}$: $x = \sqrt{3}(h-50)$ …(i)
From $\tan 60^\circ = \dfrac{h}{x}$: $x = \dfrac{h}{\sqrt{3}}$ …(ii)
From (i) and (ii): $\sqrt{3}(h-50) = \dfrac{h}{\sqrt{3}} \Rightarrow 3(h-50) = h \Rightarrow 2h = 150 \Rightarrow h = 75$ m
$$x = \frac{75}{\sqrt{3}} = 25\sqrt{3} = 25 \times 1.732 = \mathbf{43.3 \text{ m}}$$ Height of tower $= 75$ m; horizontal distance $= 43.3$ m.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower are $45^\circ$ and $60^\circ$ respectively above a 20 m high building. Find the height of the tower.
Answer
Step 1: Using $\tan 45^\circ$ for building:
$$\tan 45^\circ = \frac{20}{x} \Rightarrow x = 20 \text{ m}$$ Step 2: Using $\tan 60^\circ$ for top of tower (height from ground $= h + 20$):
$$\tan 60^\circ = \frac{h + 20}{20} \Rightarrow \sqrt{3} = \frac{h+20}{20} \Rightarrow h = 20\sqrt{3} – 20 = 20(\sqrt{3}-1)$$
$$h = 20(1.732 – 1) = 20 \times 0.732 = \mathbf{14.64 \text{ m}}$$ The height of the transmission tower is approximately 14.64 m.
From the top of a 8 m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Determine the height of the tower. (Take $\sqrt{3} = 1.732$)
Answer
In right $\triangle ABC$, $\angle B = 90^\circ$:
$$\tan 45^\circ = \frac{8}{AB} \Rightarrow AB = 8 \text{ m} \quad \text{…(i)}$$
In right $\triangle ABE$, $\angle B = 90^\circ$:
$$\tan 60^\circ = \frac{BE}{AB} \Rightarrow \sqrt{3} = \frac{h}{8} \Rightarrow h = 8\sqrt{3} \text{ m}$$
Height of tower $= BC + BE = 8 + 8\sqrt{3} = 8(1 + \sqrt{3}) = 8 \times 2.732 = \mathbf{21.856 \text{ m}}$
Case Study: A satellite flying at height $h$ is watching the top of Nanda Devi (7,816 m) and Mullayanagiri (1,930 m). The angles of depression from the satellite are $30^\circ$ and $60^\circ$ respectively. The distance between the peaks is 1937 km and the satellite is vertically above the midpoint.
Q.1 Distance of satellite from top of Nanda Devi:
(A) 1118.36 km (B) 577.52 km (C) 1937 km (D) 1025.36 km
Q.2 Distance of satellite from top of Mullayanagiri:
(A) 1139.4 km (B) 577.52 km (C) 1937 km (D) 1025.36 km
Q.3 Distance of satellite from the ground:
(A) 1139.4 km (B) 567 km (C) 1937 km (D) 1025.36 km
Q.4 Angle of elevation if a man stands 7816 m from Nanda Devi:
(A) $30^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $0^\circ$
Q.5 A milestone at $45^\circ$ to the top of Mullayanagiri. Distance of milestone:
(A) 1118.327 m (B) 566.976 m (C) 1930 m (D) 1025.36 m
Q.1 Distance of satellite from top of Nanda Devi:
(A) 1118.36 km (B) 577.52 km (C) 1937 km (D) 1025.36 km
Q.2 Distance of satellite from top of Mullayanagiri:
(A) 1139.4 km (B) 577.52 km (C) 1937 km (D) 1025.36 km
Q.3 Distance of satellite from the ground:
(A) 1139.4 km (B) 567 km (C) 1937 km (D) 1025.36 km
Q.4 Angle of elevation if a man stands 7816 m from Nanda Devi:
(A) $30^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $0^\circ$
Q.5 A milestone at $45^\circ$ to the top of Mullayanagiri. Distance of milestone:
(A) 1118.327 m (B) 566.976 m (C) 1930 m (D) 1025.36 m
Answer
Q.1 — (A) 1118.36 km
$AG = \dfrac{1937}{2}$ km. $\cos 30^\circ = \dfrac{AG}{AF} \Rightarrow AF = \dfrac{1937}{\sqrt{3}} = \dfrac{1937\sqrt{3}}{3} \approx 1118.36$ km
Q.2 — (C) 1937 km
$\cos 60^\circ = \dfrac{1937/2}{FP} \Rightarrow \dfrac{1}{2} = \dfrac{1937}{2FP} \Rightarrow FP = 1937$ km
Q.3 — (B) 567 km
$FG = \dfrac{1937}{2\sqrt{3}}$; Total $\approx \dfrac{1937}{2\sqrt{3}} + 7.816 \approx 567$ km
Q.4 — (B) $45^\circ$
$\tan\theta = \dfrac{7816}{7816} = 1 = \tan 45^\circ \Rightarrow \theta = 45^\circ$
Q.5 — (C) 1930 m
$\tan 45^\circ = \dfrac{1930}{d} \Rightarrow 1 = \dfrac{1930}{d} \Rightarrow d = 1930$ m
$AG = \dfrac{1937}{2}$ km. $\cos 30^\circ = \dfrac{AG}{AF} \Rightarrow AF = \dfrac{1937}{\sqrt{3}} = \dfrac{1937\sqrt{3}}{3} \approx 1118.36$ km
Q.2 — (C) 1937 km
$\cos 60^\circ = \dfrac{1937/2}{FP} \Rightarrow \dfrac{1}{2} = \dfrac{1937}{2FP} \Rightarrow FP = 1937$ km
Q.3 — (B) 567 km
$FG = \dfrac{1937}{2\sqrt{3}}$; Total $\approx \dfrac{1937}{2\sqrt{3}} + 7.816 \approx 567$ km
Q.4 — (B) $45^\circ$
$\tan\theta = \dfrac{7816}{7816} = 1 = \tan 45^\circ \Rightarrow \theta = 45^\circ$
Q.5 — (C) 1930 m
$\tan 45^\circ = \dfrac{1930}{d} \Rightarrow 1 = \dfrac{1930}{d} \Rightarrow d = 1930$ m
Case Study — Archery Academy: Arjuna stands 100 m from a table at point B. He shoots at balloons at height H, reducing angle by $\beta$ after the first shot. (Use $\sqrt{3} = 1.73$, $\sqrt{2} = 1.41$)
Q.1 If $\theta = 45^\circ$, $\beta = 15^\circ$, what is the difference between initial and final box height?
(A) $100 – \dfrac{100}{\sqrt{3}}$ m (B) $\dfrac{100}{\sqrt{3}}$ m (C) $100\sqrt{3} – 100$ m (D) Cannot be calculated
Q.2 Distance arrow travels to burst the second balloon ($\theta = 45^\circ$, $\beta = 15^\circ$):
(A) $\dfrac{100\sqrt{3}}{2}$ m (B) $\dfrac{200}{\sqrt{3}}$ m (C) $100\sqrt{2}$ m (D) $100\sqrt{3}$ m
Q.3 For $\theta = 60^\circ$ and $\beta = 30^\circ$, what is $\dfrac{H}{h}$?
(A) $\dfrac{1}{\sqrt{3}}$ (B) $\sqrt{3}$ (C) 2 (D) 3
Q.4 How much should Arjuna retreat to change $\theta$ from $45^\circ$ to $30^\circ$?
(A) 73 m (B) 100 m (C) 173 m (D) Cannot be calculated
Q.5 Wind pushes balloon 15 m higher when $\theta = 45^\circ$. What should Arjuna do?
(A) Move toward table 15 m, same angle. (B) Move away 15 m, same angle.
(C) Increase angle by $15^\circ$, stay in place. (D) Move away 15 m, increase angle $15^\circ$.
Q.1 If $\theta = 45^\circ$, $\beta = 15^\circ$, what is the difference between initial and final box height?
(A) $100 – \dfrac{100}{\sqrt{3}}$ m (B) $\dfrac{100}{\sqrt{3}}$ m (C) $100\sqrt{3} – 100$ m (D) Cannot be calculated
Q.2 Distance arrow travels to burst the second balloon ($\theta = 45^\circ$, $\beta = 15^\circ$):
(A) $\dfrac{100\sqrt{3}}{2}$ m (B) $\dfrac{200}{\sqrt{3}}$ m (C) $100\sqrt{2}$ m (D) $100\sqrt{3}$ m
Q.3 For $\theta = 60^\circ$ and $\beta = 30^\circ$, what is $\dfrac{H}{h}$?
(A) $\dfrac{1}{\sqrt{3}}$ (B) $\sqrt{3}$ (C) 2 (D) 3
Q.4 How much should Arjuna retreat to change $\theta$ from $45^\circ$ to $30^\circ$?
(A) 73 m (B) 100 m (C) 173 m (D) Cannot be calculated
Q.5 Wind pushes balloon 15 m higher when $\theta = 45^\circ$. What should Arjuna do?
(A) Move toward table 15 m, same angle. (B) Move away 15 m, same angle.
(C) Increase angle by $15^\circ$, stay in place. (D) Move away 15 m, increase angle $15^\circ$.
Answer
Q.1 — (A) $100 – \dfrac{100}{\sqrt{3}}$ m
At $\theta = 45^\circ$: BC $= 100$ m. At $\theta – \beta = 30^\circ$: BD $= \dfrac{100}{\sqrt{3}}$ m.
Difference $= BC – BD = 100 – \dfrac{100}{\sqrt{3}}$ m.
Q.2 — (B) $\dfrac{200}{\sqrt{3}}$ m
In $\triangle ABD$: $\sin 30^\circ = \dfrac{BD}{AD} \Rightarrow \dfrac{1}{2} = \dfrac{100/\sqrt{3}}{AD} \Rightarrow AD = \dfrac{200}{\sqrt{3}}$ m.
Q.3 — (D) 3
$H = 100\tan 60^\circ = 100\sqrt{3}$ m; $h = 100\tan 30^\circ = \dfrac{100}{\sqrt{3}}$ m.
$\dfrac{H}{h} = \dfrac{100\sqrt{3}}{\frac{100}{\sqrt{3}}} = 3$.
Q.4 — (A) 73 m
At $45^\circ$: base $= 100$ m. At $30^\circ$: base $= 100\sqrt{3} = 173$ m. Retreat $= 173 – 100 = 73$ m.
Q.5 — (B) Move away from the table by 15 m, same angle.
At $\theta = 45^\circ$, height $=$ base. The new height is $H + 15$, so Arjuna must move 15 m further away to keep height $=$ base.
At $\theta = 45^\circ$: BC $= 100$ m. At $\theta – \beta = 30^\circ$: BD $= \dfrac{100}{\sqrt{3}}$ m.
Difference $= BC – BD = 100 – \dfrac{100}{\sqrt{3}}$ m.
Q.2 — (B) $\dfrac{200}{\sqrt{3}}$ m
In $\triangle ABD$: $\sin 30^\circ = \dfrac{BD}{AD} \Rightarrow \dfrac{1}{2} = \dfrac{100/\sqrt{3}}{AD} \Rightarrow AD = \dfrac{200}{\sqrt{3}}$ m.
Q.3 — (D) 3
$H = 100\tan 60^\circ = 100\sqrt{3}$ m; $h = 100\tan 30^\circ = \dfrac{100}{\sqrt{3}}$ m.
$\dfrac{H}{h} = \dfrac{100\sqrt{3}}{\frac{100}{\sqrt{3}}} = 3$.
Q.4 — (A) 73 m
At $45^\circ$: base $= 100$ m. At $30^\circ$: base $= 100\sqrt{3} = 173$ m. Retreat $= 173 – 100 = 73$ m.
Q.5 — (B) Move away from the table by 15 m, same angle.
At $\theta = 45^\circ$, height $=$ base. The new height is $H + 15$, so Arjuna must move 15 m further away to keep height $=$ base.
Frequently Asked Questions
What is Some Applications of Trigonometry about in CBSE Class 10 Maths? ⌄
This chapter teaches students how to apply trigonometric ratios to solve real-life problems involving heights and distances. Topics include calculating the height of a tower, the width of a river, or the distance of a ship using angles of elevation and angles of depression — all without physically measuring the object.
How many marks does Some Applications of Trigonometry carry in the CBSE Class 10 board exam? ⌄
This chapter is part of the Trigonometry unit, which carries approximately 12 marks in the CBSE Class 10 board paper. Questions typically appear as 2-mark or 3-mark problems and occasionally as 5-mark case study questions, making it a chapter that offers consistent scoring opportunities.
What are the most important topics students should focus on in this chapter? ⌄
The most important topics are: finding heights using angles of elevation, finding distances using angles of depression, problems with two observers or objects on either side of a point, and case study questions involving satellites, ships, and kites. Mastering the correct trigonometric ratio (sin, cos, or tan) for each situation is key.
What common mistakes do students make in Some Applications of Trigonometry? ⌄
The most frequent error is confusing the angle of elevation with the angle of depression, or forgetting that these angles are equal when alternate interior angles are involved. Students also commonly misidentify the opposite, adjacent, or hypotenuse sides when drawing the right triangle. Regular practice with diagram-based questions helps your child avoid these mistakes.
How does Angle Belearn help students score well in Some Applications of Trigonometry? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks from real board papers, each with step-by-step solutions showing exactly how to draw the correct diagram and select the right trigonometric ratio. Regular practice with these verified questions builds both speed and accuracy so your child walks into the board exam with confidence.
