Question

From a point 100 m above a lake the angle of elevation of a stationary helicopter is ${30}^{\circ }$ and the angle of depression of reflection of the helicopter in the lake is ${60}^{\circ }$. Find the height of the helicopter above the lake.

Answer 1

Let AB be the surface of the lake and P be the point of observation such that $AP=BM=100m$.

Let C be the position of the helicopter and C' be its reflection in the lake

Then, $CB=C^{\prime }B$

Let PM be perpendicular from P on CB

Then, $\angle CPM={30}^o$ and $\angle CPM={60}^o$

Let $CM=h$

Then, $CB=h+100$ and $C^{\prime }B=h+100$

In right $△CMP$

$\Rightarrow \tan 30=\frac{CM}{PM}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{PM}$

$\Rightarrow PM=\sqrt{3}h......\left(i\right)$

In right $△PM{C}^{\prime }$

$\Rightarrow \tan 60=\frac{C^{\prime }M}{PM}$

$\Rightarrow \sqrt{3}=\frac{C^{\prime }B+BM}{PM}$

$\Rightarrow \sqrt{3}=\frac{h+100+100}{PM}$

$\Rightarrow PM=\frac{h+200}{\sqrt{3}}.....\left(ii\right)$

From (i) and (ii), we get

$\sqrt{3}h=\frac{h+200}{\sqrt{3}}$

$\Rightarrow 3h=h+200$

$\Rightarrow 2h=100$

$\Rightarrow h=100$

Now,

$CB=CM+MB=h+100=100+100=200$

Hence, the height of the helicopter from the surface of the lake is 200 m.