Two fire Tower are 34 kilometers apart, tower a being west of tower B, A fire spotted from the towers and the bearing from A and B are E 15 degree N and W 35 degree N respectively. Find the distance x of the fire from the line segment.

Answer 1

I have triangle AFB, with angle A = 15°, angle B = 35° and AB, F is where the fire is
I will assume that x is the perpendicular distance from F to line AB

Let the point of contact be P, so that x = FP

You have 2 right-angled triangles, let PB=y, then AP = 34-y
tan15 = x/(34-y) ----> x = (34-y)tan15
tan35 = x/y ---> x = ytan35

then ytan35 = (34-y)tan15
ytan35 + ytan15 = 34
y = 34/(tan35+tan15)

once you have y, sub it back into x = ytan35

Answer 2

Triangle AFB.
Given: A = 15o, B = 35o, AB = 34 km.
F = 180-(15+35) = 130o.

Law of sine:
BF/sin15 = AB/sin130
BF/sin15 = 34/sin130
BF = 34*sin15/sin130 = 11.5 km. = r
r*sin35 = 11.5 * sin35 = 6.6 km. = distance of fire from line AB.