Question

The height of the cliff can be found out if the following information is given
I. The angle of elevation of the cliff from a fixed point F is ${45}^{\circ }$
II. After going up towards a distance of 1000 m at an inclination of ${30}^{\circ }$, the angle of elevation is ${60}^{\circ }$

• A. Only statement I is required
• B. Only statement II is required
• C. Both statements I and II are required
• D. Neither of the statement is sufficient

Answer 1

The correct option is C

Both statements I and II are required

In $\Delta FPLsin{30}^{\circ }=\frac{PL}{PF}\Rightarrow PL=PFsin{30}^{\circ }=(1\times \frac{1}{2})km=\frac{1}{2}km\therefore OM=PL=\frac{1}{2}km\Rightarrow MS=OS-OM=(h-\frac{1}{2})km$
Also, $cos{30}^{\circ }=\frac{FL}{PF}\Rightarrow FL=PFcos{30}^{\circ }=(1\times \frac{\sqrt{3}}{2})=\frac{\sqrt{3}}{2}km$
Now, $h=OS=OL+LF\Rightarrow h=OL+\frac{\sqrt{3}}{2}\Rightarrow OL=(h-\frac{\sqrt{3}}{2})km\Rightarrow PM=(h-\frac{\sqrt{3}}{2})km$
In $\Delta SPM.tan{60}^{\circ }=\frac{SM}{PM}\Rightarrow SM=PMtan{60}^{\circ }\Rightarrow (h-\frac{1}{2})=(h-\frac{\sqrt{3}}{2})\sqrt{3}\Rightarrow (h-\frac{1}{2})=(\sqrt{3}n-\frac{3}{2})\Rightarrow (\sqrt{3}h-h)=\frac{3}{2}-\frac{1}{2}$
$\Rightarrow h(\sqrt{3}-1)=1\Rightarrow h=\frac{1}{\sqrt{3}-1}$
Both the statements are required