Question

A radar is observing a plane at a height of 100m and an angle of elevation of 30O. After 60s (seconds), the angle of elevation becomes 60O with the same height. What is the speed of the plane?

Answer 1

Given: A plane at a height of 100 m, when observed first is at an elevation of 30°; 60 s later, it's at an elevation of 60°.

To find: The speed of the plane.

Answer:

(Diagram for reference attached below.)

In Δ ACE,

$\tt tan\ {30}^{\circ}\ =\ \dfrac{AE}{EC}\\\\\\\dfrac{1}{\sqrt{3}}\ =\ \dfrac{100}{EC}\\\\\\EC\ =\ 100\sqrt{3}\ m$

In Δ BDC,

$\tt tan\ {60}^{\circ}\ =\ \dfrac{BD}{DC}\\\\\\\sqrt{3}\ =\ \dfrac{100}{DC}\\\\\\DC\ =\ \dfrac{100}{\sqrt{3}}\\\\\\DC\ =\ \dfrac{100\sqrt{3}}{3}\ m$

We know that:

$\tt Speed\ =\ \dfrac{Distance}{Time}$ .

We have the time (60 s). We'll need the distance, which is ED.

EC - DC = ED.

$\tt 100\sqrt{3}\ -\ \dfrac{100\sqrt{3}}{3}\ =\ ED\\\\\\\dfrac{300\sqrt{3}\ -\ 100\sqrt{3}}{3}\ =\ ED\\\\\\\dfrac{200\sqrt{3}}{3}\ m\ =\ ED$

Now, to find the speed.

$\tt Speed\ =\ \dfrac{\dfrac{200\sqrt{3}}{3}}{60}\\\\\\Speed\ =\ \dfrac{200\sqrt{3}}{3}\ \times\ \dfrac{1}{60}\\\\\\\bf Speed\ =\ \dfrac{10\sqrt{3}}{9}\ ms^-1$