Question

The angle of elevation of a jet plane from a point on the ground is 60°. After a flight of 15 seconds the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 m, find the speed of the jet plane.​

Answer 1

Note:

tan(x)=opposite/adjacent or p/b

tan(60°)=√3 and tan(30°)=1/√3

Answer:

200 m/s or 720km/hr

Step-by-step explanation:

We take 2 ∆s, initial as ABC & Final as AED.

<C=60, <D =30

BC=DE=1500√3

AB=tan(60°)*1500√3=1500*3=4500m

AE=tan(30°)*1500√3=1500√3/√3=1500m

Difference=3000m or 3km

The difference is the distance covered.

Speed=distance/time

Speed=3000/15 m/s or 3/1/240 km/hr

Speed=200m/s or 720km/hr