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 $35m$ 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 $PAQ$ be the hill, $AB$ be the tree and $PCD$ be the horizontal. Let $P$ be the point of observation.
Produce $BA$ to meet at $C$. Let $AB=Hm$.
Then, $\angle DPA=15^{\circ },PA=35m$
$\angle CPB=60^{\circ }$ and $\angle PCA=90^{\circ }$.
$\therefore \angle APB=\left(60^{\circ }-15^{\circ }\right)=45^{\circ }$
In $△PAC,\angle PAC=180^{\circ }-\left(15^{\circ }+90^{\circ }\right)=75^{\circ }$
$\therefore \angle PAB=\left(180^{\circ }-75^{\circ }\right)=105^{\circ }$
and $\angle PBA=180^{\circ }\left(45^{\circ }+105^{\circ }\right)=30^{\circ }$
Applying sine rule on $△PAB$, we get
$\frac{PA}{\sin \angle PBA}=\frac{AB}{\sin \angle APB}\Rightarrow \frac{35}{\sin 30^{\circ }}=\frac{H}{\sin 45^{\circ }}$
$\therefore 35\times 2=H\times \sqrt{2}\Rightarrow H=35\sqrt{2}m$
Hence, the height of the tree is $35\sqrt{2}m$.