Mathematics: 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?

Two fire-lookout stations are 100 miles apart,

kyriebarnes143

28.01.2022

The distance from station A to the fire is 90.4 miles

The distance from station B to the fire is 95.6 miles

Step-by-step 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.

$\frac{Sin A}{a} = \frac{Sin C}{c}$

$\frac{Sin55}{a} = \frac{Sin65}{100}$

$a = \frac{Sin55 * 100}{Sin65}$

a = 90.38

The distance from station A to the fire is 90.4 miles

For station B:

$\frac{Sin B}{b} = \frac{Sin C}{c}$

$\frac{Sin60}{b} = \frac{Sin65}{100}$

Solving for b, we have:

$a = \frac{Sin60 * 100}{Sin65}$

a = 95.6

The distance from station B to the fire is 95.6 miles

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