The angle of elevation of a jet plane from a point A on the ground is 60[^].

Question

The angle of elevation of a jet plane from a point A on the ground is 60$$^\circ$$. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30$$^\circ$$. If the jet plane is flying at a constant height, then its height is :

Accepted Answer

$$v = 432 \times {{1000} \over {60 \times 60}}$$ m/sec = 120 m/sec

Distance AB = v $$\times$$ 20 = 2400 meter

In $$\Delta$$PAC

$$\tan 60^\circ  = {h \over {PC}} \Rightarrow PC = {h \over {\sqrt 3 }}$$

In $$\Delta$$PBD

$$\tan 30^\circ  = {h \over {PD}} \Rightarrow PD = \sqrt 3 h$$

PD = PC + CD

$$\sqrt 3 h = {h \over {\sqrt 3 }} + 2400 \Rightarrow {{2h} \over {\sqrt 3 }} = 2400$$

$$h = 1200\sqrt 3 $$ meter

Solutions · 1