Question

Three deer, A, B, and C, are grazing in a field. Deer B is located 61.6 m from deer A at an angle of 52.2 ° north of west. Deer C is located 79.4 ° north of east relative to deer A. The distance between deer B and C is 91.4 m. What is the distance between deer A and C? (Hint: Consider the laws of sines and cosines given in Appendix E.)

Answer 1

Draw their positions. 61.6 degree north of west is basically you start from the negative x axis and then go up 52.2 degree. Notice B is located there. And For 79.4 degree north of east is basically you start from positive x axis and you move up 79.4 degree. But what is important is the angle of A that make with B and C. How do we do it?

180 - (79.4 + 52.2) = 48.4 degrees

A makes 48.4 degree.

length BC = 91.4 m

AB = 61.6 m

You notice that it needs more info for law of cosine.

So, now use law of sin to find angle C. Angle C is opposite side of length AB.

sinC/c = sinA/a

sinC = [(sin52.2)/91.4]*61.6

C = 32.18 degrees

Now you can find the angle B.

180 - (32.18 + 52.2) = 95.62 degree

Note that Angle B is opposite side of length AC

Now use the law of cosine.

b^2 = a^2 + c^2 - 2(a)(c)cosB

a = 91.4m

c= 61.6m

angle B = 95.62

plug the numbers in and you should get the answer.

b^2 = 91.4^2 + 61.6^2 – 2(91.4)(61.6)cos95.62

=> b = 115 m (ans)