Question

An airplane when flying at a height of 5000 m from the ground passes vertically above another airplane at an instant, when the angles of elevation of the two airplane from the same point on the ground are ${60}^{\circ }$ and ${45}^{\circ }$ respectively. The vertical distance between the airplane at that instant is:

• A. $5000(\sqrt{3}-1)m$
• B. $5000(3-\sqrt{3})m$
• C. $5000(1-\frac{1}{\sqrt{3}})m$
• D. 4500 m

Answer 1

The correct option is C $5000(1-\frac{1}{\sqrt{3}})m$

In $△BCD,tan{45}^o=\frac{5000-h}{CD}$

$\therefore CD=5000-h--------\left(i\right)$

In $△ADC,$

$tan{60}^o=\frac{5000}{CD}$

$\sqrt{3}=\frac{5000}{CD}\therefore CD=\frac{5000}{\sqrt{3}}------\left(ii\right)$

comparong eq (i) and (ii)

$5000-h=\frac{5000}{\sqrt{3}}$

$\therefore \sqrt{3}(5000-h)=5000$

$\therefore 5000\sqrt{3}-h\sqrt{3}=5000$

$5000\sqrt{3}-5000=h\sqrt{3}$

$\therefore h=\frac{5000(\sqrt{3}-1)}{\sqrt{3}}m$

$\therefore h=5000(1-\frac{1}{\sqrt{3}})m$