Question

The angles of elevation to an airplane from
two points A and B on level ground are 50◦
and 67◦
, respectively. The points A and B
are 3.9 miles apart, and the airplane is east of
both points in the same vertical plane.
Find the altitude of the plane.
Answer in units of miles. Show all work.

Answer 1

Make a sketch.
I used your A and B as points on the ground, P as the position of the plane and R as a point on the ground directly below P. I let angle A = 50°

Enter your given data.
angle ABP = 113°
Therefore angle APB = 17°

by the sine law:
PB/sin50 = 3.9/sin17
PB = 3.9sin50/sin17
= ...

In the right-angled triangle BRP
sin 67 = PR/BP
PR = BPsin 67
= 3.9 sin50 sin67/sin17
= appr 9.4

or (this is the more traditional way of doing this)

Let BR = x and PR = h
In triangle ARP:
tan 50 = h/(x+3.9)
h = (x+3.9)tan50

in triangel BRP:
tan 67 = h/x
h = xtan67

xtan67 = (x+3.9)tan50 = xtan50 + 3.9tan50
xtan67 -xtan50 = 3.9tan50
x(tan67-tan50) = 3.9tan50
x = 3.9tan50/(tan67-tan50)

back to h = xtan67
h = 3.9tan50 tan67/(tan67-tan50)
= appr 9.4 , just like above