Question

A ranger in tower A spots a fire at a direction of 302°. A ranger in tower B, located 40 mi at a direction of 41° from tower A, spots the fire at a direction of 291°. How far from tower A is the fire? How far from tower B?

The fire is __mi from tower A, and __ mi from tower B.
(round to the nearest whole number)

Answer 1

Answer on Question #41853, Math, Trigonometry

So we need to find AC and BC.

At first let's find all angles of the triangular ABC.

Angle CAB (<CAB) = 360°-<OAC + 41° = 99°

$<\mathrm{CBA}< 180{}^{\circ}-41{}^{\circ}=291-180-41=70{}^{\circ}$

$<\mathrm{ACB}< 180{}^{\circ}-<\mathrm{CAB}< 11{}^{\circ}$

Using sin theorem (http://www.cut-the-knot.org/proofs/sine_cosine.shtml):

$$\frac{AB}{\sin(<ACB)} = \frac{AC}{\sin(<ABC)} \rightarrow AC = AB \frac{\sin(<ABC)}{\sin(<ACB)} = 196.99 \, \text{mi}$$ $$\frac{AB}{\sin(<ACB)} = \frac{BC}{\sin(<CAB)} \rightarrow BC = AB \frac{\sin(<CAB)}{\sin(<ACB)} = 207.05 \, \text{mi}$$

ANSWER!!!!!!!!!!!!!!!!!!!!!

The fire is 197 mi from tower A, and 207.5 mi from tower B. (round to the nearest whole number)

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