bearing of a tower are taken from two points.A and B 250m apart along a straight north-south path, A being due north of B. the bearing of the tower from A and B are 140* and 110* respectively. what is the distance from A and B to the tower.
I see a triangle ABT, with AB = 250, angle BAT = 40° and angle ABT = 110° making angle T = 30°
By the Sine Law
From B to T
BT/sin40 = 250/sin 30
BT = 250sin40/sin30 = 321.39 m
find AT the same way.