**Q) The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 3600 √3 m, find the speed of the jet plane.**

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

**Step 1: **Let’s start from Δ APQ, tan A = tan 60° =

∴ √3 =

∴ S = 3600 m

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

∴ tan 30 =

∴

∴ S + D = 3600 √3 x √3 = 10800

∴ D = 10800 – S = 10800 – 3600 [∵ S = 3600 from part (i)]

∴ D = 7200 m

**Step 3:** Given that the jet plane covered the distance D in 30 seconds

and we calculated D = 7200 m

∴ Speed of the Jet Plane =

= meter /second

= km / hour

**= 864 km/hr**

**Therefore, speed of the jet plane is 864 km/hr**

**Please do press “Heart” button if you liked the solution.**