This page provides comprehensive Some Applications of Trigonometry – Exercise 9.1. Get step-by-step NCERT solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1. Solve problems on Heights and Distances.
Heights and Distances
Line of Sight: The line drawn from the eye of an observer to the point in the object viewed.
Angle of Elevation: Angle above the horizontal level.
Angle of Depression: Angle below the horizontal level.
1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
Let $AB$ be the pole and $AC$ be the rope.
Given: Length of rope $AC = 20$ m, $\angle C = 30^\circ$.
In $\triangle ABC$:
$\sin 30^\circ = \frac{AB}{AC}$
$\frac{1}{2} = \frac{AB}{20} \Rightarrow AB = 10$ m.
Height of the pole is 10 m.
2. 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 8 m. Find the height of the tree.
Let the tree be $AB$. It breaks at $C$, and top $A$ touches ground at $D$.
$BC$ is the standing part, $CD$ is the broken part ($CD = AC$).
Given: $BD = 8$ m, $\angle D = 30^\circ$.
1. Find $BC$: $\tan 30^\circ = \frac{BC}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{8} \Rightarrow BC = \frac{8}{\sqrt{3}}$.
2. Find $CD$: $\cos 30^\circ = \frac{BD}{CD} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{CD} \Rightarrow CD = \frac{16}{\sqrt{3}}$.
Total Height $= BC + CD = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}$ m.
Height of the tree is $8\sqrt{3}$ m.
4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Let height be $h$. Distance $d = 30$ m. Angle $\theta = 30^\circ$.
$\tan 30^\circ = \frac{h}{30}$
$\frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} = 10\sqrt{3}$ m.
Height of the tower is $10\sqrt{3}$ m.
5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Height $h = 60$ m. Angle $\theta = 60^\circ$. Let string length be $L$.
$\sin 60^\circ = \frac{60}{L}$
$\frac{\sqrt{3}}{2} = \frac{60}{L} \Rightarrow L = \frac{120}{\sqrt{3}} = 40\sqrt{3}$ m.
Length of the string is $40\sqrt{3}$ m.
6. A 1.5 m tall boy is standing at some distance from a 30 m 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 towards the building.
Effective height of building above eye level $= 30 - 1.5 = 28.5$ m.
Let initial distance be $x$ and final distance be $y$. Distance walked $= x - y$.
1. $\tan 30^\circ = \frac{28.5}{x} \Rightarrow x = 28.5\sqrt{3}$.
2. $\tan 60^\circ = \frac{28.5}{y} \Rightarrow y = \frac{28.5}{\sqrt{3}} = 9.5\sqrt{3}$.
Distance walked $= 28.5\sqrt{3} - 9.5\sqrt{3} = 19\sqrt{3}$ m.
He walked $19\sqrt{3}$ m.
9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Let Tower height $CD = 50$ m. Building height $AB = h$. Distance between feet $BD = x$.
1. In $\triangle CDB$ (Tower): $\tan 60^\circ = \frac{50}{x} \Rightarrow x = \frac{50}{\sqrt{3}}$.
2. In $\triangle ABD$ (Building): $\tan 30^\circ = \frac{h}{x} \Rightarrow h = x \tan 30^\circ$.
$h = \frac{50}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = \frac{50}{3} = 16\frac{2}{3}$ m.
Height of the building is $16.67$ m.
12. From the top of a 7 m 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 building be $AB = 7$ m. Tower be $CD$.
Draw horizontal line $AE$ from top of building to tower.
$AB = ED = 7$ m. Let $CE = h$. Total tower height $= h + 7$.
1. Depression of foot is 45° $\Rightarrow \angle EAD = 45^\circ$.
In $\triangle AED$: $\tan 45^\circ = \frac{7}{AE} \Rightarrow AE = 7$ m.
2. Elevation of top is 60° $\Rightarrow \angle CAE = 60^\circ$.
In $\triangle AEC$: $\tan 60^\circ = \frac{h}{AE} \Rightarrow h = 7\sqrt{3}$.
Total Height $= 7 + 7\sqrt{3} = 7(\sqrt{3} + 1)$ m.
Height of the tower is $7(\sqrt{3} + 1)$ m.
16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Let height be $h$. Distances are $a=4$ and $b=9$.
Angles are $\theta$ and $90-\theta$.
1. $\tan \theta = \frac{h}{4}$.
2. $\tan(90-\theta) = \frac{h}{9} \Rightarrow \cot \theta = \frac{h}{9}$.
Multiply (1) and (2):
$\tan \theta \cdot \cot \theta = \frac{h}{4} \cdot \frac{h}{9}$
$1 = \frac{h^2}{36} \Rightarrow h^2 = 36 \Rightarrow h = 6$ m.
Hence Proved.