The angle of elevation of an aircraft from a point A on the ground

Leave a Comment / Class 10th, trigonometry applications / By Sapling Academy

Q) The angle of elevation of an aircraft from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30°. The aircraft is flying at a constant height of 3500√3 m at a uniform speed. Find the speed of the aircraft.

Ans: Let’s start with the diagram for this question:

Step 1: Let’s start from Δ APQ, tan A = tan 60° = $\frac{PQ}{AQ}$

∴ √3 = $\frac{3500 \sqrt 3}{S}$

∴ S = 3500 m

Step 2: Next, we take Δ ABC, tan A = $\frac{BC}{AB}$

∴ tan 30 = $\frac{3500 \sqrt 3}{S + D}$

∴ $\frac{1}{\sqrt 3} = \frac{3500 \sqrt 3}{S + D}$

∴ S + D = 3500 √3 x √3 = 10500

∴ D = 10500 – S = 10500 – 3500

∴ D = 7000 m

Step 3: Given that the aircraft covered the distance D in 30 seconds

and we calculated D = 7000 m

∴ Speed of the Aircraft = $\frac{distance}{time}$

= $\frac{7000}{30}$ meter /second

= $\frac{700}{3} \times \frac{18}{5}$ km / hour

= 840 km/hr

Therefore, speed of the aircraft is 840 km/hr

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