Question

Two​ fire-lookout stations are 100 miles ​apart, with station A directly south of station B. Both stations spot a fire. The bearing of the fire from station A is Upper N 55 degrees Upper EN55°E and the bearing of the fire from station B is Upper S 60 degrees E.S60°E. How​ far, to the nearest tenth of a​ mile, is the fire from each lookout​ station?

Answer 1

The distance from station A to the fire is 90.4 miles

The distance from station B to the fire is 95.6 miles

Explanation:

Given that 2 fire stations are 100 miles apart.

Station A = N 55°E

Station B = S 60°E

To find the distance of the fire from each station, let's use the law of sines,

From the diagram, our values are:

A = 55°

B = 60°

Let's take C as point of fire.

Since the total sum of angle in a triangle is 180,

A+B+C=180

C = 180 - A - B

C = 180 - 55 - 60 = 65°

For station A.

`(Sin A)/(a) = (Sin C)/(c)`

`(Sin55)/(a) = (Sin65)/(100)`

`a = (Sin55 * 100)/(Sin65)`

`a = 90.38`

The distance from station A to the fire is 90.4 miles

For station B:

`(Sin B)/(b) = (Sin C)/(c)`

`(Sin60)/(b) = (Sin65)/(100)`

Solving for b, we have:

`a = (Sin60 * 100)/(Sin65)`

`a = 95.6`

The distance from station B to the fire is 95.6 miles