I drew a side-view of the situation.
Le your position on the ground be C
Let the first position of the plane be A
let the new position of the plane 2 minutes later be B
Let P be the point on the ground directly below A
given: angle PCA = 14°= angle CAB (alternate angles with parallel lines)
angle PCB = 54° which makes angle ACB = 40°
in 2 minutes, the plane went 600(2/6) = 20 miles
so AB = 20
Also angle B = 180-40-14 = 126°
by the Sine Law:
BC/sin14 = 20/sin126
BC = 5.98 miles
The height :
drop a perpendicular from B to the base CP, let it be h
We have a right-angled triangle.
Sin 54° = h/5.98
h = 5.98sin54 = 4.838 or appr 4.8 miles high
You observe a plane approaching overhead at a speed of 600 mph. The angle of elevation of the plane is 14 degrees at one time and 54 degrees two min later.
A) What is the altitude of the plane?
B) how far from the plane are you (after two minutes)?
1 answer
1 answer