Question

The angle of elevation of a jet fighter from a point A on the ground is 60°.After a fighter of 10 seconds,the angle elevation changes to 30°.If the jet is flying at a speed of 648km/h, find the constant height at which the jet is flying.

Answer 1

SOLUTION

In 3600 seconds distance travelled by a plane = 648000m.

In 10 seconds distance travelled by plane,

$= > \frac{648000}{3600} \times 10 \\ \\ = > 1800m$

In ∆ABC,

$tan60 \degree = \frac{h}{x} \\ \\ = > \sqrt{3} = \frac{h}{x} \\ \\ = > h = x \sqrt{3} .............(1)$

In ∆ADE,

$tan30 \degree = \frac{h}{x + 1800m} \\ \\ = > h = \frac{x + 1800}{ \sqrt{3} } ............(2)$

From equation (1) & (2), we get

$= > x \sqrt{3} = \frac{x + 1800}{ \sqrt{3} } \\ = > 3x = x + 1800 \\ = > 3x - x = 1800 \\ = > 2x = 1800 \\ = > x = \frac{1800}{2} = 900m \\ \\ = > x = 900m \\ \\ h = x \sqrt{3} \\ = > h = (900 \times 1.732)m \\ = > h = 1558.8m \: \: \ \: \: \: \: \: (answer)$

Hope it helps ☺️