Question

The angle of elevation of the top of a hill from a point is $\alpha$. After walking b meters towards the top up a slope inclined at an angle $\beta$ to the horizon, the angle of elevation of the top becomes $\gamma$. Then, the height of the bill is

• A. $\frac{b\sin \alpha \sin (\gamma -\beta )}{\sin (\gamma -\beta )}$
• B. $\frac{b\sin \alpha \sin (\gamma -\alpha )}{\sin (\gamma -\beta )}$
• C. $\frac{b\sin (\gamma -\alpha )}{\sin (\gamma -\beta )}$
• D. $\frac{b\sin (\gamma -\beta )}{\sin (\gamma -\alpha )}$

Answer 1

The correct option is A

$\frac{b\sin \alpha \sin (\gamma -\beta )}{\sin (\gamma -\beta )}$

Explanation for correct option:

Let BC be the hill and C be the top of hill.
$\begin{array}{rcl}\angle CAB& =& \alpha \\ PA& =& b\end{array}$
From right angle $∆PAD$

$$\begin{gathered} \sin \beta =\frac{a_2}{b} \\ \Rightarrow a_2=b\sin \beta .....\left(1\right) \end{gathered}$$

Also, from $△CQP,a_1=PC\sin \gamma ....\left(2\right)$

$$\begin{gathered} \angle PCA=\gamma -\alpha \\ \therefore \frac{b}{\sin (\gamma -\alpha )}=\frac{PC}{\sin (\alpha -\beta )} \\ \Rightarrow PC=\frac{b\sin (\alpha -\beta )}{\sin (\gamma -\alpha )} \end{gathered}$$

From $\left(2\right)$

$a_1=\frac{b\sin (\alpha -\beta )\sin \gamma }{\sin (\gamma -\alpha )}...\left(3\right)$

So, the height of the hill will be as follows.

$\begin{array}{rcl}h& =& \frac{b}{\sin (\gamma -\alpha )}\left[\sin \beta \sin \left(\gamma -\alpha \right)+\sin \left(\alpha -\beta \right)\sin (\gamma -\alpha )\sin \gamma \right]\\ & =& \frac{b\sin \alpha }{\sin (\gamma -\alpha )}\left[\sin \gamma \cos \beta -\sin \beta \cos \gamma \right]\\ & =& \frac{b\sin \alpha \sin (\gamma -\beta )}{\sin (\gamma -\beta )}\end{array}$

Hence, option (A) is correct.