Question

A tree stands vertically on a hill side which makes an angle of $15^{\circ}$ with the horizontal. From a point on the ground $35 m$ down the hill from the base of the tree, the angle of elevation of the top of the tree is $60^{\circ}$. The height of the tree is

• A. $31 \sqrt{2} m$
• B. $33 \sqrt{2} m$
• C. $35 \sqrt{2} m$
• D. $34 \sqrt{2} m$

Answer 1

Answer: (C)

Let $P A Q$ be the hill, $A B$ be the tree and $P C D$ be the horizontal. Let $P$ be the point of observation.
Produce $B A$ to meet at $C$. Let $A B=H m$.
Then, $\angle D P A=15^{\circ}, P A=35 m$
$\angle C P B=60^{\circ}$ and $\angle P C A=90^{\circ}$.
$\therefore \angle A P B=\left(60^{\circ}-15^{\circ}\right)=45^{\circ}$
In $ \triangle P A C, \angle P A C=180^{\circ}-\left(15^{\circ}+90^{\circ}\right)=75^{\circ}$
$\therefore \angle P A B=\left(180^{\circ}-75^{\circ}\right)=105^{\circ} $
and $ \angle P B A=180^{\circ} \left(45^{\circ}+105^{\circ}\right)=30^{\circ}$
Applying sine rule on $\triangle P A B$, we get
$ \frac{P A}{\sin \angle P B A}=\frac{A B}{\sin \angle A P B} \Rightarrow \frac{35}{\sin 30^{\circ}}=\frac{H}{\sin 45^{\circ}} $
$\therefore 35 \times 2=H \times \sqrt{2} \Rightarrow H=35 \sqrt{2} m$
Hence, the height of the tree is $35 \sqrt{2} m$.