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Heights and Distances PYQs

Class 10 Previous Year Questions (2016 – 2026)

Topic Overview

Some Applications of Trigonometry focuses on real-world measurements using angles of elevation and depression. Mastering the drawing of accurate diagrams is key to solving these 3-mark and 5-mark board exam staples.

Q1 2026 (SP)
00:00
The angle of elevation of the sun, when the length of the shadow of a tree is $\sqrt{3}$ times the height of the tree, is:
(a)(A) 30°
(b)(B) 45°
(c)(C) 60°
(d)(D) 90°
Let height $= h$, then shadow $= \sqrt{3}h$.
$\tan \theta = \text{height}/\text{shadow} = h/(\sqrt{3}h) = 1/\sqrt{3}$.
Since $\tan 30^{\circ} = 1/\sqrt{3}$, the angle is 30°.
Ans: (A) 30°
Q2 2025 (SP)
00:00
If the angle of elevation of the top of a tower from a point 30m away from its foot is 45°, then the height of the tower is:
(a)(A) 15m
(b)(B) 30m
(c)(C) $30\sqrt{3}$m
(d)(D) 45m
Distance (base) $= 30$m, $\theta = 45^{\circ}$.
$\tan 45^{\circ} = \text{height}/\text{base} \Rightarrow 1 = h/30 \Rightarrow h = 30$m.
Ans: (B) 30m
Q3 2024
00:00
A ladder makes an angle of 60° with the ground. If the foot of the ladder is 2.5m away from the wall, the length of the ladder is:
(a)(A) 5m
(b)(B) 2.5m
(c)(C) $5\sqrt{3}$m
(d)(D) 4m
Base $= 2.5$m, $\theta = 60^{\circ}$. We need Hypotenuse (ladder length).
$\cos 60^{\circ} = \text{Base}/\text{Hypotenuse} \Rightarrow 1/2 = 2.5/L \Rightarrow L = 5$m.
Ans: (A) 5m
Q4 2023
00:00
Assertion (A): If the shadow of a pole increases, the angle of elevation of the sun also increases.
Reason (R): As the sun moves lower in the sky, shadows become longer.
As the sun moves lower (angle $\theta$ decreases), the shadow ($h \cot \theta$) increases.
Thus, if shadow increases, $\theta$ must be DECREASING, not increasing.
Assertion is False; Reason is True.
Ans: (D) A is false, R is true
Q5 2022
00:00
The angle of depression of a car from the top of a 75m high tower is 30°. The distance of the car from the base of the tower is:
(a)(A) 25m
(b)(B) $25\sqrt{3}$m
(c)(C) $75\sqrt{3}$m
(d)(D) 150m
Height $= 75$m, $\theta = 30^{\circ}$.
$\tan 30^{\circ} = 75/d \Rightarrow 1/\sqrt{3} = 75/d \Rightarrow d = 75\sqrt{3}$m.
Ans: (C) $75\sqrt{3}$m
Q6 2021
00:00
If a pole 6m high casts a shadow $2\sqrt{3}$m long on the ground, then find the sun's elevation.
$\tan \theta = \text{Height}/\text{Shadow} = 6/(2\sqrt{3}) = 3/\sqrt{3} = \sqrt{3}$.
Since $\tan 60^{\circ} = \sqrt{3}$, the elevation is 60°.
Ans: 60°
Q7 2020
00:00
The ratio of the length of a vertical rod and its shadow is $1 : \sqrt{3}$. Find the angle of elevation of the sun.
$\tan \theta = 1/\sqrt{3} \Rightarrow \theta = 30^{\circ}$.
Ans: 30°
Q8 2024
00:00
A bridge over a river makes an angle of 45° with the river bank. If the length of the bridge across the river is 150m, what is the width of the river?
Let width of river $= w$. Length of bridge (hypotenuse) $= 150$m.
$\sin 45^{\circ} = w/150 \Rightarrow 1/\sqrt{2} = w/150$.
$w = 150/\sqrt{2} = 75\sqrt{2}$m.
Ans: $75\sqrt{2}$m
Q9 2023
00:00
From a point on the ground, 20m away from the foot of a vertical tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower.
$h/20 = \tan 60^{\circ} = \sqrt{3}$.
$h = 20\sqrt{3}$m.
Ans: $20\sqrt{3}$m
Q10 2019
00:00
The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 30m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.
Let height $= h$, initial distance $= x+30$.
$h/(x+30) = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x+30 = h\sqrt{3}$.
$h/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow x = h/\sqrt{3}$.
Substitute $x$: $h/\sqrt{3} + 30 = h\sqrt{3} \Rightarrow 30 = h(\sqrt{3} - 1/\sqrt{3}) = h(2/\sqrt{3})$.
$h = 15\sqrt{3}$m.
Ans: $15\sqrt{3}$m
Q26 2016
00:00
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree.
Let $x$ be the standing part and $y$ be the broken part.
$\tan 30^{\circ} = x/8 \Rightarrow 1/\sqrt{3} = x/8 \Rightarrow x = 8/\sqrt{3}$.
$\cos 30^{\circ} = 8/y \Rightarrow \sqrt{3}/2 = 8/y \Rightarrow y = 16/\sqrt{3}$.
Total height $= x+y = 24/\sqrt{3} = 8\sqrt{3}$m.
Ans: $8\sqrt{3}$m
Q27 2021
00:00
A statue, 1.6m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and the angle of elevation of the top of the pedestal is 45°. Find height of pedestal.
Let pedestal $= h$. Dist to point $= x$.
$h/x = \tan 45^{\circ} = 1 \Rightarrow h = x$.
$(h+1.6)/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h+1.6 = h\sqrt{3}$.
$h = 1.6/(\sqrt{3}-1) = 0.8(\sqrt{3}+1)$m.
Ans: $0.8(\sqrt{3}+1)$m
Q11 2024
00:00
A 1.5m tall boy is standing at some distance from a 30m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked.
Effective height $= 30 - 1.5 = 28.5$m.
Let distance walked be $d$. Let final distance from building be $x$.
$28.5/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow x = 28.5/\sqrt{3} = 9.5\sqrt{3}$m.
$28.5/(x+d) = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x+d = 28.5\sqrt{3}$.
$d = 28.5\sqrt{3} - 9.5\sqrt{3} = 19\sqrt{3}$m.
Ans: $19\sqrt{3}$m
Q12 2023
00:00
As observed from the top of a 75m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the ships.
Height $= 75$m.
For 45° ship: $75/x_1 = \tan 45^{\circ} = 1 \Rightarrow x_1 = 75$m.
For 30° ship: $75/x_2 = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x_2 = 75\sqrt{3}$m.
Distance between ships $= 75\sqrt{3} - 75 = 75(\sqrt{3}-1)$m.
Ans: $75(\sqrt{3}-1)$m
Q13 2022
00:00
From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Let tower height $= H$. Building height $= 7$m.
Angle of depression to foot $= 45^{\circ} \Rightarrow$ Distance between building and tower $= 7/\tan 45^{\circ} = 7$m.
In upper triangle: $(H-7)/7 = \tan 60^{\circ} = \sqrt{3}$.
$H-7 = 7\sqrt{3} \Rightarrow H = 7(\sqrt{3}+1)$m.
Ans: $7(\sqrt{3}+1)$m
Q14 2020
00:00
Two poles of equal heights are standing opposite each other on either side of a road 80m wide. From a point between them on the road, angles of elevation are 60° and 30°. Find height of poles.
Let height $= h$. Distances from point $= x$ and $80-x$.
$h/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h = x\sqrt{3}$.
$h/(80-x) = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow h\sqrt{3} = 80-x$.
Substitute $h$: $(x\sqrt{3})\sqrt{3} = 80-x \Rightarrow 3x = 80-x \Rightarrow 4x = 80 \Rightarrow x = 20$.
$h = 20\sqrt{3}$m.
Ans: $20\sqrt{3}$m
Q15 2018
00:00
The angle of elevation of the top of a tower from two points at a distance of 4m and 9m from base are complementary. Prove height is 6m.
Let height $= h$. Angles are $\theta$ and $90-\theta$.
$h/4 = \tan \theta$
$h/9 = \tan(90-\theta) = \cot \theta$.
Multiply: $(h/4) \times (h/9) = \tan \theta \times \cot \theta = 1$.
$h^2/36 = 1 \Rightarrow h = 6$m.
Ans: Height is 6m
Q28 2018
00:00
A man on the deck of a ship, 10m above water level, observes elevation of hill top as 60° and depression of base as 30°. Find distance and hill height.
Let distance $= x$. Angle of depression $30^{\circ} \Rightarrow 10/x = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x = 10\sqrt{3}$m.
Let hill height above deck be $h'$. $h'/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h' = 10\sqrt{3} \times \sqrt{3} = 30$m.
Total height $= 30 + 10 = 40$m.
Ans: Dist = $10\sqrt{3}$m, Height = 40m
Q31 2024
00:00
A tower is surmounted by a 5m flagstaff. From a point, elevations of bottom and top of flagstaff are 30° and 60°. Find tower height.
Let tower $= h$, dist $= x$. $h/x = \tan 30^{\circ} = 1/\sqrt{3}$.
$(h+5)/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h+5 = (h\sqrt{3})\sqrt{3} = 3h$.
$2h = 5 \Rightarrow h = 2.5$m.
Ans: 2.5m
Q34 2024 (SP)
00:00
The angle of elevation of building top from foot of a 60m pole is 30°; pole top from building foot is 60°. Find building height.
Let building $= h$, dist $= x$. $60/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow x = 20\sqrt{3}$m.
$h/x = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow h = 20\sqrt{3}/\sqrt{3} = 20$m.
Ans: 20m
Q16 2024
00:00
The angle of elevation of a cloud from a point 60m above a lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud.
Let height of cloud above lake $= H$. Point is 60m above lake.
Height of cloud from observation level $= H - 60$.
Distance to shadow reflection from observation level $= H + 60$.
Let horizontal distance $= x$.
$(H-60)/x = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x = (H-60)\sqrt{3}$.
$(H+60)/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow H+60 = x\sqrt{3}$.
Substitute $x$: $H+60 = (H-60)\sqrt{3} \times \sqrt{3} = 3H - 180$.
$2H = 240 \Rightarrow H = 120$m.
Ans: 120m
Q17 2023
00:00
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower.
Building height $= 20$m. Let tower height $= h$.
Distance from point to building $= 20/\tan 45^{\circ} = 20$m.
Total height $(h+20)/20 = \tan 60^{\circ} = \sqrt{3}$.
$h+20 = 20\sqrt{3} \Rightarrow h = 20(\sqrt{3}-1)$m.
Ans: $20(\sqrt{3}-1)$m
Q18 2022
00:00
A straight highway leads to the foot of a tower. A man at the top observes a car at 30° depression. 6 seconds later, depression is 60°. Find time to reach tower.
Let height $= h$, initial distance $= x_1$, final distance $= x_2$.
$h/x_1 = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x_1 = h\sqrt{3}$.
$h/x_2 = \tan 60^{\circ} = \sqrt{3} \Rightarrow x_2 = h/\sqrt{3}$.
Distance covered in 6s $= x_1 - x_2 = h\sqrt{3} - h/\sqrt{3} = h(2/\sqrt{3})$.
Speed $= \text{Distance}/\text{Time} = [h(2/\sqrt{3})]/6 = h/(3\sqrt{3})$.
Time for $x_2 = x_2/\text{Speed} = (h/\sqrt{3}) / (h/3\sqrt{3}) = 3$ seconds.
Ans: 3 seconds
Q19 2020
00:00
The angle of elevation of the top of a building from the foot of a tower is 30° and elevation of top of tower from foot of building is 60°. If tower is 50m high, find building height.
Let tower $= 50$m, building $= h$. Distance $= x$.
$50/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow x = 50/\sqrt{3}$.
$h/x = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow h = x/\sqrt{3}$.
$h = (50/\sqrt{3}) / \sqrt{3} = 50/3 = 16.67$m.
Ans: 16.67m
Q20 2017
00:00
From the top of a 60m high building, the angles of depression of top and bottom of a tower are 45° and 60° respectively. Find height of tower.
Building $= 60$m. Let tower $= h$.
Distance to tower $= 60/\tan 60^{\circ} = 60/\sqrt{3} = 20\sqrt{3}$m.
Drop in building height $(60-h) = \text{Dist} \times \tan 45^{\circ} = 20\sqrt{3} \times 1$.
$h = 60 - 20\sqrt{3} = 20(3-\sqrt{3})$m.
Ans: $20(3-\sqrt{3})$m
Q29 2015
00:00
Angle of elevation of plane is 60°. After 15s flight, elevation is 30°. Plane speed is 720 km/h. Find height.
Speed $= 720$ km/h $= 200$ m/s. Dist in 15s $= 200 \times 15 = 3000$m.
Let height $= h$. $h/x_1 = \sqrt{3} \Rightarrow x_1 = h/\sqrt{3}$.
$h/(x_1+3000) = 1/\sqrt{3} \Rightarrow h\sqrt{3} = h/\sqrt{3} + 3000$.
$h(3-1)/\sqrt{3} = 3000 \Rightarrow h = 1500\sqrt{3}$m.
Ans: $1500\sqrt{3}$m
Q30 2020
00:00
Elevation of jet is 60°. After 15s, it is 30°. Height is $1500\sqrt{3}$m. Find speed in km/h.
$h = 1500\sqrt{3}$. $x_1 = h/\tan 60^{\circ} = 1500$m.
$x_2 = h/\tan 30^{\circ} = 4500$m.
Dist $= 4500-1500 = 3000$m.
Speed $= 3000/15 = 200$ m/s $= 720$ km/h.
Ans: 720 km/h
Q33 2017
00:00
Bird on 80m high tree. Elevation 45°. Flies horizontally. After 2s, elevation 30°. Find speed.
$h = 80$. $x_1 = 80/\tan 45^{\circ} = 80$m.
$x_2 = 80/\tan 30^{\circ} = 80\sqrt{3}$m.
Dist $= 80(\sqrt{3}-1) = 80(0.732) = 58.56$m.
Speed $= 58.56/2 = 29.28$ m/s.
Ans: 29.28 m/s
Q35 2026 (SP)
00:00
Elevation of cloud from 20m above lake is 30°; depression of reflection is 60°. Find cloud height.
Let cloud $= H$. $(H-20)/x = 1/\sqrt{3}, (H+20)/x = \sqrt{3}$.
$H+20 = 3(H-20) = 3H - 60 \Rightarrow 2H = 80 \Rightarrow H = 40$m.
Ans: 40m
Q21 2026 (SP)
00:00
PASSAGE: A light house is a tower designed to emit light from a system of lamps and lenses to serve as a navigational aid. Two ships are sailing on the same side. Angles of depression are 30° and 60° from the 100m high light house.
(i) Draw the diagram and label angles.
(ii) Find distance of Ship 1 (60°) from the light house.
(iii) (a) Find distance between ships. OR (b) If ships move apart such that angles become 30° and 45°, find new distance.
(i) Triangle with 100m height.
(ii) $d_1 = 100/\tan 60^{\circ} = 100/\sqrt{3}$m.
(iii) (a) $d_2 = 100/\tan 30^{\circ} = 100\sqrt{3}$m. Dist $= 100\sqrt{3} - 100/\sqrt{3} = 200/\sqrt{3}$m.
Q22 2025 (SP)
00:00
PASSAGE: Kites were used in ancient China for various purposes. A boy is flying a kite at a height of 75m with a string length of 150m.
(i) Find the angle of elevation of the kite.
(ii) If the string is loosened and the angle becomes 30°, what is the new height for same string length?
(iii) (a) Find horizontal distance in Case 1. OR (b) If angle increases to 60°, find string length for 75m height.
(i) $\sin \theta = 75/150 = 1/2 \Rightarrow \theta = 30^{\circ}$.
(ii) New height $h = 150 \sin 30^{\circ} = 75$m (Already same? Question implies change).
(iii) (b) $L = 75/\sin 60^{\circ} = 75/(\sqrt{3}/2) = 50\sqrt{3}$m.
Q23 2024
00:00
PASSAGE: Flagstaff on Building. A 5m flagstaff is on top of a building. From point P on ground, elevation of top of building is 45° and top of flagstaff is 60°.
(i) Find the distance of P from the building.
(ii) Find height of the building.
(iii) (a) Find total height. OR (b) If point P moves away, what happens to elevation?
(i/ii) Let building $= h$, dist $= x$. $h/x = \tan 45^{\circ} = 1 \Rightarrow h = x$.
Total height $(h+5)/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h+5 = h\sqrt{3} \Rightarrow h(\sqrt{3}-1) = 5$.
$h = 5/(\sqrt{3}-1) = 2.5(\sqrt{3}+1)$m. Dist $= h$.
Q24 2023
00:00
PASSAGE: Observation Plane. A pilot at 3000m altitude sees two ships at 30° and 45° depression on opposite sides of the plane.
(i) Find horizontal distance of Ship 1.
(ii) Find horizontal distance of Ship 2.
(iii) (a) Find total distance between ships. OR (b) Find distance if on same side.
(i) $d_1 = 3000/\tan 45^{\circ} = 3000$m.
(ii) $d_2 = 3000/\tan 30^{\circ} = 3000\sqrt{3}$m.
(iii) (a) Dist $= 3000(\sqrt{3}+1)$m.
Q25 2022
00:00
PASSAGE: Hot Air Balloon. Elevation from ground is 60°. After 10 mins flight vertically up, elevation is 30°. Dist from base is 100m.
(i) Initial height of balloon.
(ii) Final height of balloon.
(iii) (a) Distance covered in 10 mins. OR (b) Average speed in m/min.
(i) $h_1 = 100 \tan 60^{\circ} = 100\sqrt{3}$m.
(ii) Wait, if it moves UP, elevation should increase if base is fixed? Question likely meant it moved AWAY. Assuming it moved UP: $h_2 = 100 \tan 75^{\circ}$? If elevation decreased to 30°, it moved AWAY.
Assuming it moved away: Dist 1 $= h/\sqrt{3}$, Dist 2 $= h\sqrt{3}$.
00:00
The angle of elevation of the sun, when the length of the shadow of a tree is $\sqrt{3}$ times the height of the tree, is:
(a)(A) 30°
(b)(B) 45°
(c)(C) 60°
(d)(D) 90°
Let height $= h$, then shadow $= \sqrt{3}h$.
$\tan \theta = \text{height}/\text{shadow} = h/(\sqrt{3}h) = 1/\sqrt{3}$.
Since $\tan 30^{\circ} = 1/\sqrt{3}$, the angle is 30°.
Ans: (A) 30°
Q2 2025 (SP)
00:00
If the angle of elevation of the top of a tower from a point 30m away from its foot is 45°, then the height of the tower is:
(a)(A) 15m
(b)(B) 30m
(c)(C) $30\sqrt{3}$m
(d)(D) 45m
Distance (base) $= 30$m, $\theta = 45^{\circ}$.
$\tan 45^{\circ} = \text{height}/\text{base} \Rightarrow 1 = h/30 \Rightarrow h = 30$m.
Ans: (B) 30m
Q3 2024
00:00
A ladder makes an angle of 60° with the ground. If the foot of the ladder is 2.5m away from the wall, the length of the ladder is:
(a)(A) 5m
(b)(B) 2.5m
(c)(C) $5\sqrt{3}$m
(d)(D) 4m
Base $= 2.5$m, $\theta = 60^{\circ}$. We need Hypotenuse (ladder length).
$\cos 60^{\circ} = \text{Base}/\text{Hypotenuse} \Rightarrow 1/2 = 2.5/L \Rightarrow L = 5$m.
Ans: (A) 5m
Q4 2023
00:00
Assertion (A): If the shadow of a pole increases, the angle of elevation of the sun also increases.
Reason (R): As the sun moves lower in the sky, shadows become longer.
As the sun moves lower (angle $\theta$ decreases), the shadow ($h \cot \theta$) increases.
Thus, if shadow increases, $\theta$ must be DECREASING, not increasing.
Assertion is False; Reason is True.
Ans: (D) A is false, R is true
Q5 2022
00:00
The angle of depression of a car from the top of a 75m high tower is 30°. The distance of the car from the base of the tower is:
(a)(A) 25m
(b)(B) $25\sqrt{3}$m
(c)(C) $75\sqrt{3}$m
(d)(D) 150m
Height $= 75$m, $\theta = 30^{\circ}$.
$\tan 30^{\circ} = 75/d \Rightarrow 1/\sqrt{3} = 75/d \Rightarrow d = 75\sqrt{3}$m.
Ans: (C) $75\sqrt{3}$m
Q6 2021
00:00
If a pole 6m high casts a shadow $2\sqrt{3}$m long on the ground, then find the sun's elevation.
$\tan \theta = \text{Height}/\text{Shadow} = 6/(2\sqrt{3}) = 3/\sqrt{3} = \sqrt{3}$.
Since $\tan 60^{\circ} = \sqrt{3}$, the elevation is 60°.
Ans: 60°
Q7 2020
00:00
The ratio of the length of a vertical rod and its shadow is $1 : \sqrt{3}$. Find the angle of elevation of the sun.
$\tan \theta = 1/\sqrt{3} \Rightarrow \theta = 30^{\circ}$.
Ans: 30°
Q8 2024
00:00
A bridge over a river makes an angle of 45° with the river bank. If the length of the bridge across the river is 150m, what is the width of the river?
Let width of river $= w$. Length of bridge (hypotenuse) $= 150$m.
$\sin 45^{\circ} = w/150 \Rightarrow 1/\sqrt{2} = w/150$.
$w = 150/\sqrt{2} = 75\sqrt{2}$m.
Ans: $75\sqrt{2}$m
Q9 2023
00:00
From a point on the ground, 20m away from the foot of a vertical tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower.
$h/20 = \tan 60^{\circ} = \sqrt{3}$.
$h = 20\sqrt{3}$m.
Ans: $20\sqrt{3}$m
Q10 2019
00:00
The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 30m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.
Let height $= h$, initial distance $= x+30$.
$h/(x+30) = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x+30 = h\sqrt{3}$.
$h/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow x = h/\sqrt{3}$.
Substitute $x$: $h/\sqrt{3} + 30 = h\sqrt{3} \Rightarrow 30 = h(\sqrt{3} - 1/\sqrt{3}) = h(2/\sqrt{3})$.
$h = 15\sqrt{3}$m.
Ans: $15\sqrt{3}$m
Q26 2016
00:00
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree.
Let $x$ be the standing part and $y$ be the broken part.
$\tan 30^{\circ} = x/8 \Rightarrow 1/\sqrt{3} = x/8 \Rightarrow x = 8/\sqrt{3}$.
$\cos 30^{\circ} = 8/y \Rightarrow \sqrt{3}/2 = 8/y \Rightarrow y = 16/\sqrt{3}$.
Total height $= x+y = 24/\sqrt{3} = 8\sqrt{3}$m.
Ans: $8\sqrt{3}$m
Q27 2021
00:00
A statue, 1.6m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and the angle of elevation of the top of the pedestal is 45°. Find height of pedestal.
Let pedestal $= h$. Dist to point $= x$.
$h/x = \tan 45^{\circ} = 1 \Rightarrow h = x$.
$(h+1.6)/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h+1.6 = h\sqrt{3}$.
$h = 1.6/(\sqrt{3}-1) = 0.8(\sqrt{3}+1)$m.
Ans: $0.8(\sqrt{3}+1)$m
Q11 2024
00:00
A 1.5m tall boy is standing at some distance from a 30m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked.
Effective height $= 30 - 1.5 = 28.5$m.
Let distance walked be $d$. Let final distance from building be $x$.
$28.5/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow x = 28.5/\sqrt{3} = 9.5\sqrt{3}$m.
$28.5/(x+d) = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x+d = 28.5\sqrt{3}$.
$d = 28.5\sqrt{3} - 9.5\sqrt{3} = 19\sqrt{3}$m.
Ans: $19\sqrt{3}$m
Q12 2023
00:00
As observed from the top of a 75m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the ships.
Height $= 75$m.
For 45° ship: $75/x_1 = \tan 45^{\circ} = 1 \Rightarrow x_1 = 75$m.
For 30° ship: $75/x_2 = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x_2 = 75\sqrt{3}$m.
Distance between ships $= 75\sqrt{3} - 75 = 75(\sqrt{3}-1)$m.
Ans: $75(\sqrt{3}-1)$m
Q13 2022
00:00
From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Let tower height $= H$. Building height $= 7$m.
Angle of depression to foot $= 45^{\circ} \Rightarrow$ Distance between building and tower $= 7/\tan 45^{\circ} = 7$m.
In upper triangle: $(H-7)/7 = \tan 60^{\circ} = \sqrt{3}$.
$H-7 = 7\sqrt{3} \Rightarrow H = 7(\sqrt{3}+1)$m.
Ans: $7(\sqrt{3}+1)$m
Q14 2020
00:00
Two poles of equal heights are standing opposite each other on either side of a road 80m wide. From a point between them on the road, angles of elevation are 60° and 30°. Find height of poles.
Let height $= h$. Distances from point $= x$ and $80-x$.
$h/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h = x\sqrt{3}$.
$h/(80-x) = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow h\sqrt{3} = 80-x$.
Substitute $h$: $(x\sqrt{3})\sqrt{3} = 80-x \Rightarrow 3x = 80-x \Rightarrow 4x = 80 \Rightarrow x = 20$.
$h = 20\sqrt{3}$m.
Ans: $20\sqrt{3}$m
Q15 2018
00:00
The angle of elevation of the top of a tower from two points at a distance of 4m and 9m from base are complementary. Prove height is 6m.
Let height $= h$. Angles are $\theta$ and $90-\theta$.
$h/4 = \tan \theta$
$h/9 = \tan(90-\theta) = \cot \theta$.
Multiply: $(h/4) \times (h/9) = \tan \theta \times \cot \theta = 1$.
$h^2/36 = 1 \Rightarrow h = 6$m.
Ans: Height is 6m
Q28 2018
00:00
A man on the deck of a ship, 10m above water level, observes elevation of hill top as 60° and depression of base as 30°. Find distance and hill height.
Let distance $= x$. Angle of depression $30^{\circ} \Rightarrow 10/x = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x = 10\sqrt{3}$m.
Let hill height above deck be $h'$. $h'/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h' = 10\sqrt{3} \times \sqrt{3} = 30$m.
Total height $= 30 + 10 = 40$m.
Ans: Dist = $10\sqrt{3}$m, Height = 40m
Q31 2024
00:00
A tower is surmounted by a 5m flagstaff. From a point, elevations of bottom and top of flagstaff are 30° and 60°. Find tower height.
Let tower $= h$, dist $= x$. $h/x = \tan 30^{\circ} = 1/\sqrt{3}$.
$(h+5)/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h+5 = (h\sqrt{3})\sqrt{3} = 3h$.
$2h = 5 \Rightarrow h = 2.5$m.
Ans: 2.5m
Q34 2024 (SP)
00:00
The angle of elevation of building top from foot of a 60m pole is 30°; pole top from building foot is 60°. Find building height.
Let building $= h$, dist $= x$. $60/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow x = 20\sqrt{3}$m.
$h/x = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow h = 20\sqrt{3}/\sqrt{3} = 20$m.
Ans: 20m
Q16 2024
00:00
The angle of elevation of a cloud from a point 60m above a lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud.
Let height of cloud above lake $= H$. Point is 60m above lake.
Height of cloud from observation level $= H - 60$.
Distance to shadow reflection from observation level $= H + 60$.
Let horizontal distance $= x$.
$(H-60)/x = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x = (H-60)\sqrt{3}$.
$(H+60)/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow H+60 = x\sqrt{3}$.
Substitute $x$: $H+60 = (H-60)\sqrt{3} \times \sqrt{3} = 3H - 180$.
$2H = 240 \Rightarrow H = 120$m.
Ans: 120m
Q17 2023
00:00
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower.
Building height $= 20$m. Let tower height $= h$.
Distance from point to building $= 20/\tan 45^{\circ} = 20$m.
Total height $(h+20)/20 = \tan 60^{\circ} = \sqrt{3}$.
$h+20 = 20\sqrt{3} \Rightarrow h = 20(\sqrt{3}-1)$m.
Ans: $20(\sqrt{3}-1)$m
Q18 2022
00:00
A straight highway leads to the foot of a tower. A man at the top observes a car at 30° depression. 6 seconds later, depression is 60°. Find time to reach tower.
Let height $= h$, initial distance $= x_1$, final distance $= x_2$.
$h/x_1 = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow x_1 = h\sqrt{3}$.
$h/x_2 = \tan 60^{\circ} = \sqrt{3} \Rightarrow x_2 = h/\sqrt{3}$.
Distance covered in 6s $= x_1 - x_2 = h\sqrt{3} - h/\sqrt{3} = h(2/\sqrt{3})$.
Speed $= \text{Distance}/\text{Time} = [h(2/\sqrt{3})]/6 = h/(3\sqrt{3})$.
Time for $x_2 = x_2/\text{Speed} = (h/\sqrt{3}) / (h/3\sqrt{3}) = 3$ seconds.
Ans: 3 seconds
Q19 2020
00:00
The angle of elevation of the top of a building from the foot of a tower is 30° and elevation of top of tower from foot of building is 60°. If tower is 50m high, find building height.
Let tower $= 50$m, building $= h$. Distance $= x$.
$50/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow x = 50/\sqrt{3}$.
$h/x = \tan 30^{\circ} = 1/\sqrt{3} \Rightarrow h = x/\sqrt{3}$.
$h = (50/\sqrt{3}) / \sqrt{3} = 50/3 = 16.67$m.
Ans: 16.67m
Q20 2017
00:00
From the top of a 60m high building, the angles of depression of top and bottom of a tower are 45° and 60° respectively. Find height of tower.
Building $= 60$m. Let tower $= h$.
Distance to tower $= 60/\tan 60^{\circ} = 60/\sqrt{3} = 20\sqrt{3}$m.
Drop in building height $(60-h) = \text{Dist} \times \tan 45^{\circ} = 20\sqrt{3} \times 1$.
$h = 60 - 20\sqrt{3} = 20(3-\sqrt{3})$m.
Ans: $20(3-\sqrt{3})$m
Q29 2015
00:00
Angle of elevation of plane is 60°. After 15s flight, elevation is 30°. Plane speed is 720 km/h. Find height.
Speed $= 720$ km/h $= 200$ m/s. Dist in 15s $= 200 \times 15 = 3000$m.
Let height $= h$. $h/x_1 = \sqrt{3} \Rightarrow x_1 = h/\sqrt{3}$.
$h/(x_1+3000) = 1/\sqrt{3} \Rightarrow h\sqrt{3} = h/\sqrt{3} + 3000$.
$h(3-1)/\sqrt{3} = 3000 \Rightarrow h = 1500\sqrt{3}$m.
Ans: $1500\sqrt{3}$m
Q30 2020
00:00
Elevation of jet is 60°. After 15s, it is 30°. Height is $1500\sqrt{3}$m. Find speed in km/h.
$h = 1500\sqrt{3}$. $x_1 = h/\tan 60^{\circ} = 1500$m.
$x_2 = h/\tan 30^{\circ} = 4500$m.
Dist $= 4500-1500 = 3000$m.
Speed $= 3000/15 = 200$ m/s $= 720$ km/h.
Ans: 720 km/h
Q33 2017
00:00
Bird on 80m high tree. Elevation 45°. Flies horizontally. After 2s, elevation 30°. Find speed.
$h = 80$. $x_1 = 80/\tan 45^{\circ} = 80$m.
$x_2 = 80/\tan 30^{\circ} = 80\sqrt{3}$m.
Dist $= 80(\sqrt{3}-1) = 80(0.732) = 58.56$m.
Speed $= 58.56/2 = 29.28$ m/s.
Ans: 29.28 m/s
Q35 2026 (SP)
00:00
Elevation of cloud from 20m above lake is 30°; depression of reflection is 60°. Find cloud height.
Let cloud $= H$. $(H-20)/x = 1/\sqrt{3}, (H+20)/x = \sqrt{3}$.
$H+20 = 3(H-20) = 3H - 60 \Rightarrow 2H = 80 \Rightarrow H = 40$m.
Ans: 40m
Q21 2026 (SP)
00:00
PASSAGE: A light house is a tower designed to emit light from a system of lamps and lenses to serve as a navigational aid. Two ships are sailing on the same side. Angles of depression are 30° and 60° from the 100m high light house.
(i) Draw the diagram and label angles.
(ii) Find distance of Ship 1 (60°) from the light house.
(iii) (a) Find distance between ships. OR (b) If ships move apart such that angles become 30° and 45°, find new distance.
(i) Triangle with 100m height.
(ii) $d_1 = 100/\tan 60^{\circ} = 100/\sqrt{3}$m.
(iii) (a) $d_2 = 100/\tan 30^{\circ} = 100\sqrt{3}$m. Dist $= 100\sqrt{3} - 100/\sqrt{3} = 200/\sqrt{3}$m.
Q22 2025 (SP)
00:00
PASSAGE: Kites were used in ancient China for various purposes. A boy is flying a kite at a height of 75m with a string length of 150m.
(i) Find the angle of elevation of the kite.
(ii) If the string is loosened and the angle becomes 30°, what is the new height for same string length?
(iii) (a) Find horizontal distance in Case 1. OR (b) If angle increases to 60°, find string length for 75m height.
(i) $\sin \theta = 75/150 = 1/2 \Rightarrow \theta = 30^{\circ}$.
(ii) New height $h = 150 \sin 30^{\circ} = 75$m (Already same? Question implies change).
(iii) (b) $L = 75/\sin 60^{\circ} = 75/(\sqrt{3}/2) = 50\sqrt{3}$m.
Q23 2024
00:00
PASSAGE: Flagstaff on Building. A 5m flagstaff is on top of a building. From point P on ground, elevation of top of building is 45° and top of flagstaff is 60°.
(i) Find the distance of P from the building.
(ii) Find height of the building.
(iii) (a) Find total height. OR (b) If point P moves away, what happens to elevation?
(i/ii) Let building $= h$, dist $= x$. $h/x = \tan 45^{\circ} = 1 \Rightarrow h = x$.
Total height $(h+5)/x = \tan 60^{\circ} = \sqrt{3} \Rightarrow h+5 = h\sqrt{3} \Rightarrow h(\sqrt{3}-1) = 5$.
$h = 5/(\sqrt{3}-1) = 2.5(\sqrt{3}+1)$m. Dist $= h$.
Q24 2023
00:00
PASSAGE: Observation Plane. A pilot at 3000m altitude sees two ships at 30° and 45° depression on opposite sides of the plane.
(i) Find horizontal distance of Ship 1.
(ii) Find horizontal distance of Ship 2.
(iii) (a) Find total distance between ships. OR (b) Find distance if on same side.
(i) $d_1 = 3000/\tan 45^{\circ} = 3000$m.
(ii) $d_2 = 3000/\tan 30^{\circ} = 3000\sqrt{3}$m.
(iii) (a) Dist $= 3000(\sqrt{3}+1)$m.
Q25 2022
00:00
PASSAGE: Hot Air Balloon. Elevation from ground is 60°. After 10 mins flight vertically up, elevation is 30°. Dist from base is 100m.
(i) Initial height of balloon.
(ii) Final height of balloon.
(iii) (a) Distance covered in 10 mins. OR (b) Average speed in m/min.
(i) $h_1 = 100 \tan 60^{\circ} = 100\sqrt{3}$m.
(ii) Wait, if it moves UP, elevation should increase if base is fixed? Question likely meant it moved AWAY. Assuming it moved UP: $h_2 = 100 \tan 75^{\circ}$? If elevation decreased to 30°, it moved AWAY.
Assuming it moved away: Dist 1 $= h/\sqrt{3}$, Dist 2 $= h\sqrt{3}$.