Question

An aeroplane is flying horizontally directly towards the city at an altitude of 400 metres. At a given time the pilot views the city lights of Melbourne at an angle of depression of 1.5 degrees. Two minutes later the angle of depression of the city lights is 5 degrees. Find the speed of the aeroplane in km/hr correct to one decimal.

Answer 1

I drew a horizontal flight path 400 m above the ground.
On that path, I labeled the first position of the plane as A and the 2nd position as B, and the location of Melbourne on the ground as M
I also labeled C on the flight path as the point directly above M.
Let the speed of the plane be v km/h
then AB = (1/30)v or v/30
In triangle ABM, angle AMB = 3.5°

by the sine law:
BM/sin 1.5° = (v/30) /sin 3.5°
BM = (v/30)sin 1.5° / sin 3.5

in the right-angled triangle BCM,
sin 5° = .4/BM , (400 m = .4 km)
BM = .4/sin 5

(v/30)sin 1.5° / sin 3.5 = .4/sin 5
v/30 = (.4/sin5)(sin3.5/sin1.5)
v = 30(.4/sin5)(sin3.5/sin1.5)
= .....

Answer 2

A diagram and review of the basic trig functions will show that the distance d flown by the plane between observations is

d = 400 cot 1.5° - 400 cot 5° = 10700m = 10.7km
10.7km/2min * 60min/hr = 321 km/hr