Question

$A$ tree stands vertical, on the hill side, which makes an angle of ${22}^0$ with the horizontal. From the point $35$ meters directly down the hill from the base of the tree, the angle of elevation of the top of the tree is ${45}^0$. Then the height of the tree (Given $\sin {22}^0=0.3746,\cos {22}^{}=0.9276$ from tables) is

• A. $18.4$ m
• B. $20.4$ m
• C. $18.54$ m
• D. $30$ m

Answer 1

The correct option is C $18.54$ m

Let $PQ$ be tree and $A$ be the point on ground $AR$ such that $AD=35m$

Now $\angle QAR={45}^0$ and $\angle PAR={22}^0$

$\Rightarrow \angle PAQ=\angle QAR-\angle PAR={45}^0-{22}^0={23}^0$

$\Rightarrow \angle AQP={90}^0-\angle QAR={90}^0-{45}^0={45}^0$

Using sine rule, we get

$\frac{AP}{\sin {45}^0}=\frac{PQ}{\sin {23}^0}\Rightarrow \frac{35}{\frac{1}{\sqrt{2}}}=\frac{PQ}{0.3746}\Rightarrow PQ=18.54$