Question

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.)

Answer 1

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