An aeroplane flying horizontally at a height of 2500m above the ground is observed at a certain point on earth - Maths - Some Applications of Trigonometry

Question

An aeroplane flying horizontally at a height of 2500m
above the ground is observed at a certain point on earth to subtend
and angle of 60 0 After 15 seconds, its
angle of elevation at the same point is observed to be 30
0 Calculate the speed of the aeroplane

Priyanka Kedia · Accepted Answer

In âˆ†ABC, ${\cot 60}^{0} = \frac{A C}{B C} \Rightarrow \frac{1}{\sqrt{3}} = \frac{A C}{2500} \Rightarrow A C = \frac{2500}{\sqrt{3}} (1)$ In âˆ†AED, ${\cot 30}^{0} = \frac{A E}{E D} \Rightarrow \sqrt{3} = \frac{A E}{2500} \Rightarrow A E = 2500 \sqrt{3} (2)$ From figure, BD = CE and, CE = AE - AC so, BD = AE - AC now, substitute value of AE and AC from equation (1) and (2), we get, $B D = 2500 \sqrt{3} - \frac{2500}{\sqrt{3}} = 2500 (\sqrt{3} - \frac{1}{\sqrt{3}}) = 2500 \times \frac{2}{\sqrt{3}}$ $\Rightarrow B D = \frac{5000}{\sqrt{3}} m$ BD is covered by aeroplane in 15 seconds So, $S p e e d o f t h e a e r o p l a n e = \frac{D i s \tan c e (B D)}{T i m e (t)} = \frac{\frac{5000}{\sqrt{3}}}{15} = \frac{5000}{15 \times 1 732} = 192 455 m / s$ Hence, speed of the aeroplane is 192 455 m/s

An aeroplane flying horizontally at a height of 2500m above the ground is observed at a certain point on earth to subtend and angle of 60 0 . After 15 seconds, its angle of elevation at the same point is observed to be 30 0 . Calculate the speed of the aeroplane.

An aeroplane flying horizontally at a height of 2500m above the ground is observed at a certain point on earth to subtend and angle of 60 ⁰ . After 15 seconds, its angle of elevation at the same point is observed to be 30 ⁰ . Calculate the speed of the aeroplane.