Vardaan Learning Institute • ICSE Class 10 Mathematics • Chapter 22
Line of Sight: The line drawn from the eye of an observer to the point in the object viewed by the observer.
Angle of Elevation: The angle formed by the line of sight with the horizontal line when the object is above the horizontal level of the observer's eye.
Angle of Depression: The angle formed by the line of sight with the horizontal line when the object is below the horizontal level of the observer's eye.
Golden Rule for Angle of Depression: Angle of depression of object $B$ as seen from observer $A$ is equal to the angle of elevation of $A$ as seen from $B$ (Alternate Interior Angles).
$$\text{Angle of Depression} = \text{Angle of Elevation}$$
| Angle ($\theta$) | $30^\circ$ | $45^\circ$ | $60^\circ$ |
|---|---|---|---|
| $\sin \theta$ | $\frac{1}{2} = 0.5$ | $\frac{1}{\sqrt{2}} \approx 0.7071$ | $\frac{\sqrt{3}}{2} \approx 0.8660$ |
| $\cos \theta$ | $\frac{\sqrt{3}}{2} \approx 0.8660$ | $\frac{1}{\sqrt{2}} \approx 0.7071$ | $\frac{1}{2} = 0.5$ |
| $\tan \theta$ | $\frac{1}{\sqrt{3}} \approx 0.5774$ | $1$ | $\sqrt{3} \approx 1.732$ |
Standard Constants: $\sqrt{2} = 1.414$, $\sqrt{3} = 1.732$
Type 1: Two Ships on Opposite Sides of a Lighthouse
Problem: From the top of a $100\text{ m}$ high lighthouse, the angles of depression of two ships on opposite sides of the lighthouse are $30^\circ$ and $45^\circ$. Find the distance between the two ships.
Solution:
Let $AB = 100\text{ m}$ be the height of the lighthouse. Let $C$ and $D$ be the two ships on opposite sides.
In right $\triangle ABC$: $\tan 45^\circ = \frac{AB}{BC} \implies 1 = \frac{100}{BC} \implies BC = 100\text{ m}$
In right $\triangle ABD$: $\tan 30^\circ = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{100}{BD} \implies BD = 100\sqrt{3}\text{ m}$
Distance between ships $CD = BC + BD = 100 + 100\sqrt{3} = 100(1 + 1.732) = 100 \times 2.732 = 273.2\text{ m}$.
Type 2: Observer Moving Towards a Tower
Problem: A man standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is $60^\circ$. When he moves $40\text{ m}$ away from the bank, the angle of elevation becomes $30^\circ$. Find the height of the tree and the width of the river.
Solution:
Let $h$ be height of tree and $x$ be width of river.
In $\triangle 1$: $\tan 60^\circ = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3}$
In $\triangle 2$: $\tan 30^\circ = \frac{h}{x + 40} \implies \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 40}$
$$x + 40 = 3x \implies 2x = 40 \implies x = 20\text{ m}$$
Height of tree $h = 20\sqrt{3} = 20 \times 1.732 = 34.64\text{ m}$.
BOARD From the top of a cliff $60\text{ m}$ high, the angles of depression of the top and bottom of a tower are observed to be $30^\circ$ and $60^\circ$ respectively. Find the height of the tower.
BOARD An aeroplane at an altitude of $250\text{ m}$ observes two ships sailing towards it in the same direction. The angles of depression of the ships are $45^\circ$ and $60^\circ$. Find the distance between the two ships.