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sukhopar [10]
3 years ago
15

Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer

c is located 79.1 ° north of east relative to deer
a. the distance between deer b and c is 93.8 m. what is the distance between deer a and c? (hint: consider the laws of sines and cosines given in appendix
e.)
Physics
1 answer:
sveticcg [70]3 years ago
4 0

by cosine law we know that

c^2 = a^2 + b^2 - 2 abcos\theta

\theta = 180 - 54.3 - 79.1 = 46.6 degree

now using above equation

93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6

4942.03 = b^2 - 85.34 b

b^2 - 85.34b - 4942.03 = 0

by solving above quadratic equation we have

b = 124.9 m

so it is at distance 124.9 m from deer a

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