Question

From a point on a hill-side of constant inclination, the angle of elevation of the top of a flagstaff on its summit is observed to be $\alpha$ and $a$ meter nearer the top of the hill, it is $\beta$. If $h$ is the height of the flagstaff, the inclination of the hill to the horizon is

• A. ${\sin }^{-1}\frac{\left[a\sin \alpha \sin \beta \right]}{\left[h\sin \right(\beta -\alpha \left)\right]}$
• B. $co{s}^{-1}\frac{\left[a\sin \alpha \sin \beta \right]}{\left[h\sin \right(\beta -\alpha \left)\right]}$
• C. $ta{n}^{-1}\frac{\left[a\sin \alpha \sin \beta \right]}{\left[h\sin \right(\beta -\alpha \left)\right]}$
• D. none of these

Answer 1

The correct option is B

$co{s}^{-1}\frac{\left[a\sin \alpha \sin \beta \right]}{\left[h\sin \right(\beta -\alpha \left)\right]}$

Step 1: Draw a rough diagram from the given information.

Let OP be the height of the flagstaff given as $h$.

$\begin{array}{rcl}\angle ORP& =& \alpha ,\\ \angle PQO& =& \beta ,\\ \angle OAB& =& \theta ,\\ \angle RPQ& =& \beta -\alpha ,\\ \angle POQ& =& \frac{\mathrm{\pi }}{2}+\theta \\ RQ& =& a\end{array}$

Step 2: Apply sine's law.

In$△PQR$,

$$\begin{gathered} \frac{PR}{\sin (\pi -\beta )}=\frac{RQ}{\sin (\beta -\alpha )};\{\frac{a}{\sin A}=\frac{b}{\sin B}\} \\ \Rightarrow \frac{PR}{\sin \left(\beta \right)}=\frac{a}{\sin (\beta -\alpha )} \\ \Rightarrow PR=\frac{a\sin \left(\beta \right)}{\sin (\beta -\alpha )} \end{gathered}$$

In$△POR$,

$$\begin{gathered} \frac{PR}{\sin ({\displaystyle \frac{\mathrm{\pi }}{2}}+\theta )}=\frac{PO}{\sin \left(\alpha \right)};\{\frac{a}{\sin A}=\frac{b}{\sin B}\} \\ \Rightarrow \frac{PR}{\cos \left(\theta \right)}=\frac{h}{\sin \left(\alpha \right)} \\ \Rightarrow PR=\frac{h\cos \left(\theta \right)}{\sin \left(\alpha \right)} \end{gathered}$$

from above,

$$\begin{gathered} \frac{a\sin \left(\beta \right)}{\sin (\beta -\alpha )}=\frac{h\cos \left(\theta \right)}{\sin \left(\alpha \right)} \\ \Rightarrow \cos \left(\theta \right)=\frac{a\sin \left(\alpha \right)\sin \left(\beta \right)}{h\sin (\beta -\alpha )} \\ \Rightarrow \theta ={\cos }^{-1}\left[\frac{a\sin \left(\alpha \right)\sin \left(\beta \right)}{h\sin (\beta -\alpha )}\right] \end{gathered}$$

Hence, option$\mathbf{}\mathbf{\left(}\mathit{B}\mathbf{\right)}$ is the correct option.