Question

A man observes the elevation of a balloon to be ${30}^0.$ He then walks 1 km. towards the balloon and finds the angle of elevation now is ${60}^0.$ The height (in km) of the balloon is

• A. $\frac{\sqrt{3}}{2}$
• B. $\sqrt{3}+1$
• C. $\sqrt{3}-1$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

Answer: (A)

$\frac{\sqrt{3}}{2}$

Consider triangle ABCD

Let AB=h be the height of the balloon

Let CD be the distance walked by the man

It is given that

$\angle D={30}^0$ and $\angle C={60}^0$ which are the angles of elevation before and after the man travelled a distance of 1Km

Consider triangle ABC

$\tan {60}^0=\frac{AB}{BC}=\sqrt{3}$

$\frac{AB}{\sqrt{3}}=BC$ ...(i)

Consider triangle ABD

$\tan {30}^0=\frac{1}{\sqrt{3}}$

$=\frac{AB}{1+BC}$

$1+BC=\sqrt{3}AB$

$1=\sqrt{3}AB-\frac{AB}{\sqrt{3}}$ ... from i

Therefore $\frac{2AB}{\sqrt{3}}=1$

$AB=\frac{\sqrt{3}}{2}$

Hence answer is A