JEE Main 2018 — Heights and Distances Question with Solution

Let the height of the tower $MN$ be $h$

In $∆QMN$

$\tan 30°=\frac{MN}{QM}$

$\therefore QM=\sqrt{3} h=MR \dots \left(1\right)\{\because M$ is the point $QM=MR\}$

Now in $\Delta MNP$ 

$\because \angle MPN=\angle NMP=45°$

$\therefore MN=PM(\because$ In a triangle sides opposite to equal angles are

equal$...\left(2\right)$ )

$\therefore MN=PM=h$

In $∆PMQ$

By using Equation $\left(1\right)$

$PM=\sqrt{(200{)}^2-(\sqrt{3} h{)}^2}$ {By using Pythagoras theorem}

$\therefore$ From $\left(2\right)$

$h=\sqrt{(200{)}^2-(\sqrt{3}h{)}^2}$

$\Rightarrow h=\sqrt{\left(40000\right)-3{ h}^2}$

$\Rightarrow h^2=40000-3{ h}^2$ {By squaring both the sides}

$\Rightarrow 4{ h}^2=40000$

$\Rightarrow h^2=10000 \mathrm{m}$

$\Rightarrow h=100 \mathrm{m}$