Gm Pls answer the following question - Maths - Some Applications of Trigonometry

Question

Gm Pls answer the following question

Rahul K · Accepted Answer

Dear Student,

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 Regards

Gm...Pls answer the following question...