Posted by **Mike** on Tuesday, May 10, 2011 at 2:25pm.

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.

- Math -
**Henry**, Wednesday, May 11, 2011 at 4:37pm
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.

- Math -
**sam**, Wednesday, December 4, 2013 at 5:50am
21.9

- Math -
**Arpita**, Thursday, November 5, 2015 at 2:02am
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

- Math -
**Arpita**, Thursday, November 5, 2015 at 2:05am
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

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