Math
posted by Mike .
Two fire towers are 30 kilometers apart, tower A being due west of tower B. A fire is spotted from the towers, and the bearings from A and B are E 14 degrees N and W 34 degrees N, respectively. Find the distance d of the fire from the line segment AB.

1. Draw line segment AB.
2. Draw the 14degree angle from point
A.
3. Draw the 34degree angle from point B. Label the intersection of these 2 lines point C. Now we have formed
triangle ABC.
4. Draw the altitude from point C perpendicular to AB and label it CD.
The distance of the fire from AB is
equal to the altitude(CD).
A + B + C = 180 Deg.
14 + 34 + C = 180,
C = 132 Deg.
a/sinA = c/sinC,
a/sin14 = 30/sin132,
Multiply both sides by sin14:
a = 30sin14 / sin132 = 9.77km.
CD = 9.77sin34 = 5.46km = dist. from
fire to AB. 
21.9

1. Call the point where d intersects AB point C.
2. Let CB equal x.
3. cot(14)= (30x)/d
cot(34)= x/d
4. cot(14)= (30/d) (x/d)
cot(14)= (30/d) cot(34)
cot(14)+ cot(34)= (30/d)
d(cot14+ cot34)= 30
d = 30/ (cot14+ cot34)
d = 5.46 km 
Sorry this one's easier to read.
1..
Call the point where d intersects AB point C.
2..
Let CB equal x.
3..
cot(14)= (30x)/d
cot(34)= x/d
4..
cot(14)= (30/d) (x/d)
cot(14)= (30/d) cot(34)
cot(14)+ cot(34)= (30/d)
d(cot14+ cot34)= 30
d = 30/ (cot14+ cot34)
d = 5.46 km