**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° =

∴ √3 =

∴ S = 3500 m

**Step 2: **Next, we take Δ ABC, tan A =

∴ tan 30 =

∴

∴ 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 =

= meter /second

= km / hour

**= 840 km/hr**

**Therefore, speed of the aircraft is 840 km/hr**

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