Question

A ranger in tower A spots a fire at a direction of 312 degrees. A ranger in tower​ B, located 40 mi at a direction of 54 degrees from tower​ A, spots the fire at a direction of 272 degrees. How far from tower A is the​ fire? How far from tower​ B?

Answer 1

From tower A: `38.31" mi"`
From tower B: `60.87" mi"`

Explanation:

The following is a drawing of the sightings for `color(red)"tower A"` and `color(blue)"tower B"`:

The fire is at the intersection of the two rays.

Let `angle A =` angle between the first tower and second tower

`angle A = 360^@-312^@+54^@`

`angle A = 102^@`

Let `angle C =` the angle that between the two towers at the fire

`angle C = 312^@-272^@`

`angle C = 40^@`

We shall find the measure of `angle B` using the properties of a triangle:

`180^@= 102^@+40^@ + angleB`

`angleB = 38^@`

We know the that the length of side `c = 40" mi"`

We can use the law of sines to find the length of side b (the distance from tower A to the fire):

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

`b = sin(38^@)(40" mi")/sin(40^@)`

`b = 38.31" mi"`

We can use the law of sines to find the length of side a (the distance from tower B to the fire):

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

`a = sin(102^@)(40" mi")/sin(40^@)`

`b = 60.87" mi"`