Questions: A ranger in tower A spots a fire at a direction of 347°. A ranger in tower B, located 60 mi at a direction of 59° from tower A, spots the fire at a direction of 317°. How far from tower A is the fire? How far from tower B? The fire is from tower A. (Round to the nearest whole number.) The fire is from tower B.

Solution Steps

We need to find the distances from two towers (A and B) to a fire. The directions from the towers to the fire are given, and the distance between the towers is known.

• Tower A spots the fire at a direction of 347°.
• Tower B spots the fire at a direction of 59°.
• The distance between Tower A and Tower B is 40 miles.
• The angle between the directions from the towers to the fire is 317° - 59° = 258°.

We can use the Law of Sines to find the distances from the towers to the fire. The Law of Sines states: $$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$$ where $a$, $b$, and $c$ are the sides of the triangle, and $A$, $B$, and $C$ are the opposite angles.

• Angle at Tower A (A) = 360° - 347° = 13°
• Angle at Tower B (B) = 59°
• Angle at the fire (C) = 180° - (13° + 59°) = 108°

Using the Law of Sines: $$\frac{AB}{\sin(C)} = \frac{a}{\sin(A)} = \frac{b}{\sin(B)}$$ where $AB = 40$ miles, $C = 108°$, $A = 13°$, and $B = 59°$.

$$\frac{40}{\sin(108°)} = \frac{a}{\sin(13°)} = \frac{b}{\sin(59°)}$$

Solving for $a$ (distance from Tower A to the fire): $$a = \frac{40 \cdot \sin(13°)}{\sin(108°)}$$

Solving for $b$ (distance from Tower B to the fire): $$b = \frac{40 \cdot \sin(59°)}{\sin(108°)}$$

Final Answer