Question

Two towers labeled Tower A and Tower B some distance apart and a fire image at a certain distance from both towers. Towers A and B are located 10 miles apart. A ranger spots a fire at a 42-degree angle from tower A. Another fire ranger spots the same fire at a 64-degree angle from tower B. To the nearest tenth of a mile, how far from tower A is the fire?

a. 7.0 mi
b. 7.4 mi
c. 9.4 mi
d. 13.4 mi

Answer 1

Using the law of sines with the given angles, 42 degrees from Tower A and 64 degrees from Tower B, and the distance between the towers, which is 10 miles, we can calculate that the fire is approximately 7.4 miles from Tower A.

Step-by-step explanation:

To find out how far from Tower A the fire is, we can use triangle trigonometry. Since the rangers at Towers A and B both spot the same fire, we can assume the towers and the fire's location form a triangle. Considering Tower A, the fire, and the line between Tower A and B as a triangle (with Tower A and B being 10 miles apart), we can determine the distance from Tower A to the fire using the law of sines.

The law of sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle. If we label the distance from Tower A to the fire as 'a' and the distance from Tower B to the fire as 'b', we can set up the following equations using the angles given (42 degrees for Tower A and 64 degrees for Tower B):

a/sin(42 degrees) = 10/sin(64 degrees)

Solving for 'a' we get:

a = 10 * sin(42 degrees) / sin(64 degrees)

After performing the calculation, we find that 'a' is approximately 7.4 miles. Therefore, the correct answer is b. 7.4 mi.