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.

Answer 1

1. Draw line segment AB.

2. Draw the 14-degree angle from point
A.

3. Draw the 34-degree 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.

Answer 2

21.9

Answer 3

1. Call the point where d intersects AB point C.
2. Let CB equal x.
3. cot(14)= (30-x)/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

Answer 4

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)= (30-x)/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