Chapter 9

Class 10 Mathematics - Chapter 9: Some Applications of Trigonometry
Chapter 9: Some Applications of Trigonometry
1. Key Terminology
1. Line of Sight: The line drawn from the eye of an observer to the point in the object viewed by the observer.

2. Angle of Elevation: The angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.

3. Angle of Depression: The angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level.
Observer (A) Object (C) B θ (Elevation) Line of Sight
2. Single-Triangle Problems
Strategy: Identify the right-angled triangle, locate the known side and angle, and apply the appropriate trigonometric ratio ($\sin \theta$, $\cos \theta$, or $\tan \theta$) to find the unknown side or angle.
Example 2.1: Height of a Tower

A tower stands vertically on the ground. From a point on the ground, which is $15\text{ m}$ away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^\circ$. Find the height of the tower.

Solution:
Let $AB$ be the height of the tower ($h$) and $BC = 15\text{ m}$ be the distance from the foot.
In right $\triangle ABC$, right-angled at $B$: $$\tan 60^\circ = \frac{AB}{BC}$$ $$\sqrt{3} = \frac{h}{15}$$ $$h = 15\sqrt{3}\text{ m}$$ The height of the tower is $15\sqrt{3}\text{ m}$ (or $\approx 25.98\text{ m}$).

Example 2.2: Length of a String / Hypotenuse

A kite is flying at a height of $60\text{ m}$ 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 $60^\circ$. Find the length of the string, assuming that there is no slack in the string.

Solution:
Let $AB = 60\text{ m}$ be the height and $AC = l$ be the length of the string.
In right $\triangle ABC$: $$\sin 60^\circ = \frac{AB}{AC}$$ $$\frac{\sqrt{3}}{2} = \frac{60}{l}$$ $$l \cdot \sqrt{3} = 120$$ $$l = \frac{120}{\sqrt{3}}$$ Rationalising the denominator: $$l = \frac{120 \times \sqrt{3}}{3} = 40\sqrt{3}\text{ m}$$ The length of the string is $40\sqrt{3}\text{ m}$.

3. Two-Triangle Problems
In complex scenarios involving two triangles (e.g., observing from a building top or moving observer), set up simultaneous equations using $\tan \theta_1$ and $\tan \theta_2$ to eliminate common variables.
Example 3.1: Two Positions / Change in Angle

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^\circ$ and $45^\circ$, respectively. If the bridge is at a height of $3\text{ m}$ from the banks, find the width of the river.

Solution:
Let $P$ be the point on the bridge, $A$ and $B$ be the two banks. $PD = 3\text{ m}$ is the perpendicular height.

In right $\triangle PDA$ with angle $30^\circ$: $$\tan 30^\circ = \frac{PD}{AD}$$ $$\frac{1}{\sqrt{3}} = \frac{3}{AD}$$ $$AD = 3\sqrt{3}\text{ m}$$
In right $\triangle PDB$ with angle $45^\circ$: $$\tan 45^\circ = \frac{PD}{DB}$$ $$1 = \frac{3}{DB}$$ $$DB = 3\text{ m}$$
Width of the river ($AB$): $$AB = AD + DB$$ $$AB = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1)\text{ m}$$ The width of the river is $3(\sqrt{3} + 1)\text{ m}$.