The angle of elevation of a jet fighter from point A on the ground is 60Â After a flight - Maths - Some Applications of Trigonometry

Question

The angle of elevation of a jet fighter from point A on the
ground is 60Â° After a flight of 15 seconds, the angle of
elevation changes to 30Â° If the jet is flying at a speed
of 720 km/h, find the constant height at which jet is flying

Ajanta Trivedi · Accepted Answer

let A be the observation point and let B and C be the two positions of jet fighter it takes 15 sec to reach the point C from B let the height of the jet fighter from the ground be BE = CD = h km the speed of the jet = 720 km/hr =720000m/hr distance = speed*time therefore BC = $720000 * \frac{15}{60 * 60} = 3000 m = 3 k m$ in the right angled triangle BEA, $\tan 60 = \frac{B E}{A E} \sqrt{3} = \frac{h}{A E} \Rightarrow A E = \frac{h}{\sqrt{3}} k m (1)$ in the right angled triangle CDA, $\tan 30 = \frac{C D}{A D} \frac{1}{\sqrt{3}} = \frac{h}{A D} \Rightarrow A D = \sqrt{3} h k m (2)$ AD = AE+ED = AE+BC $\sqrt{3} h = \frac{h}{\sqrt{3}} + 3 \sqrt{3} h - \frac{h}{\sqrt{3}} = 3 \frac{3 h - h}{\sqrt{3}} = 3 \frac{2 h}{\sqrt{3}} = 3 \Rightarrow h = \frac{3}{2} \sqrt{3} k m = \frac{3}{2} * 1 732 = 2 598 k m$ thus the height of the jet fighter is 2 598 km hope this helps you

The angle of elevation of a jet fighter from point A on the ground is 60°.After a flight of 15 seconds, the angle of elevation changes to 30°.If the jet is flying at a speed of 720 km/h, find the constant height at which jet is flying.