The angle of elevation from a point on a ground to the airplane is 60°. After 15 seconds angle of elevation of the plane become 30°. If the plane is flying at the height of 1500√3 meters then the speed of the plane is?

Question

Question

Answer 1

22.10.2018.0070

AB = DE = 1500√3

IN Δ ABC

\(tan\;60^\circ = \frac{{AB}}{{BC}}\)

\(BC = \frac{{1500\sqrt 3 }}{{\sqrt 3 }} = 1500\)

In Δ DEC

\(tan\;30^\circ = \frac{{DE}}{{CE}}\)

\(\frac{1}{{\sqrt 3 }} = \frac{{1500\sqrt 3 }}{{{\rm{CE}}}}\)

CE = 1500 × 3 = 4500

BE = CE – BC

BE = 4500 – 1500

BE = 3000

Speed \( = \frac{{{\rm{distance}}}}{{{\rm{time}}}} = \frac{{3000}}{{15}} = 200\)