Question

Three deer, A, B, and C, are grazing in a field. Deer B is located 63.4 m from deer A at an angle of 52.4 ° north of west. Deer C is located 77.3 ° north of east relative to deer A. The distance between deer B and C is 94.8 m. What is the distance between deer A and C?

Answer 1

Answer:

The distance between deer A and C is 122 m

Explanation:

Hi there!

For this problem, we can use the law of sines. Please, see the attached figure for a better understanding of the problem.

From the law of sines:

sin β / side BC = sin θ/ side AB = sin α / side AC

The angle β can be calculated (see figure):

β = 180° - 52.4° - 77.3° = 50.3°

Then, we can find the angle θ :

sin β / side BC = sin θ/ side AB

sin 50.3° / 94.8 m = sin θ / 63.4 m

63.4 m · (sin 50.3° / 94.8 m) = sin θ

θ = 31.0°

Now, we can calculate the angle α. The sum of the internal angles of a triangle is 180°. Then:

α = 180° -31.0° - 50.3° = 98.7°

Using the laws of sines:

sin α / side AC = sin θ / side AB

sin 98.7°/ side AC = sin 31.0°/ 63.4 m

side AC = sin 98.7 / ( sin 31.0°/ 63.4 m)

side AC = 122 m

The distance between deer A and C is 122 m