IIT-JEE 1993Question 1
An observer at $O$ notices that the angle of elevation of the top of a tower is $30^\circ$. The line joining $O$ to the base of the tower makes an angle of $\tan^{-1}\left(1/\sqrt2\right)$ with the North and is inclined Eastwards. The observer travels a distance of $300$ meters towards the North to a point $A$ and finds the tower to his East. The angle of elevation of the top of the tower at $A$ is $\phi$. Find $\phi$ and the height of the tower.
Trigonometric Ratio & Identities
Answer$\phi=45^\circ$, height $=150\sqrt2$ m
Solution
Concept. Split $OP$ into its north and east parts. At $A$ the tower is due east, so the north part of $OP$ is $300$ m and the east part is $AP$. Formulas used. (1) $\tan\theta=\frac1{\sqrt2}\Rightarrow\cos\theta=\sqrt{\tfrac23}$, $\sin\theta=\frac{1}{\sqrt3}$(2) Height $=$ distance $\times\tan(\text{elevation})$Step 1 - Top view. Let $P$ be the base and $OP=d$.Step 2 - Find $d$. By (1), $d\cos\theta=300$, so $d=300\sqrt{\tfrac32}=150\sqrt6$. Step 3 - Height. By (2), $h=d\tan30^\circ=\dfrac{150\sqrt6}{\sqrt3}=150\sqrt2$ m. Step 4 - Elevation at $A$. $AP=d\sin\theta=\dfrac{150\sqrt6}{\sqrt3}=150\sqrt2$, so $\tan\phi=\dfrac{h}{AP}=1$ and $\phi=45^\circ$. Answer: $\phi=45^\circ$, height $=150\sqrt2$ m