Question

A forest fire is spotted from two fire towers.
The triangle formed by the two towers and the fire
has angles of 29° at tower A and 37° at tower B. If
the towers are 3000 meters apart, how far is each
tower from the fire?

Answer 1

Answer:

11,346 m and 8,346 m

Step-by-step explanation:

The situation is represented in the figure attached. We have:

- O is the position of the fire

- A' is the position of tower A on the ground

- B' is the position of tower B on the ground

$29^{\circ}$ is the angle formed by the triangle AOA'

$37^{\circ}$ is the angle formed by the triangle BOB'

The distance between A' and B' is A'B' = 3000 m

Using trigonometry, we have the following relationships:

$tan 29^{\circ} = \frac{AA'}{A'O}$

$tan 37^{\circ} = \frac{BB'}{B'O}$

They can be rewritten as

$A'O = \frac{AA'}{tan 29^{\circ}}\\B'O = \frac{BB'}{tan 37^{\circ}}$

We also know that

$A'O-B'O=3000$

Substittuing the two previous equations into this ,we get

$\frac{AA'}{tan 29^{\circ}}-\frac{BB'}{tan 37^{\circ}}=3000$

Assuming that the two towers are identical, they have same height, so

$AA'=BB'=h$

And we can solve the equation to find the height of the towers:

$\frac{h}{tan 29^{\circ}}-\frac{h}{tan 37^{\circ}}=3000\\h=\frac{3000}{\frac{1}{tan 29^{\circ}}-\frac{1}{tan 37^{\circ}}}=6289 m$

And so, the distance of each tower from the fire is:

$A'O'=\frac{h}{tan 29^{\circ}}=11,346 m\\B'O'=\frac{h}{tan 37^{\circ}}=8,346 m$