let the plane's position be P
so in triangle ABP, angle A = 30° ( a bearing of 120 = 90+30)
let's find angle P
sinP/250 = sin 30/200
(case 1)
angle P = 38.682
then angle A = 113.328°
BP/sin113.328 = 200/sin30
BP = 372.63 km
case 2
angle P = 141.318
then angle A = 8.682°
BP/sin8.682 = 200/sin30
BP = 60.38 km
both answers make sense, since the triangle results in the "ambiguous case"
An airplane leaves airport A and flies 200 km. At this time its bearing from Airport B, 250 km to the west, is 120 degrees. How far is the airplane from B?
2 answers
1. A passenger in an airplane at an altitude of 10 km sees two towns to the east of the plane. The angle of depression to town A is 28o and the angle of depression to town B is 55o. How far apart are the two towns?