Question

The horizontal distance between two towers is $60m$ and angular depression of the top of the first as seen from the second, which is $150m$ in height, is ${30}^0$. The height of the first tower is

• A. $(150+20\sqrt{3})m$
• B. $(150+15\sqrt{3})m$
• C. $(150-20\sqrt{5})m$
• D. $(150-20\sqrt{3})m$

Answer 1

The correct option is D $(150-20\sqrt{3})m$

Let AB and CD be the two towers

Height of the second tower, AB =150m and

Distance between two towers BD =60m

Let's assume the height of the first tower is $hm$.

We know, In right angled $△$,

$\tan \left(\Theta \right)=\frac{Opposite}{Adjacent}$

$\tan {30}^0=\frac{AE}{EC}$

$\tan {30}^0=\frac{150-h}{60}$ .....[AE=AB-ED] and [EC=BD=60]

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{150-h}{60}$

$\Rightarrow h\sqrt{3}=150\sqrt{3}-60$

$\Rightarrow h=150-20\sqrt{3}$

The height of the first tower is $150-20\sqrt{3}$ m.