Three deer, A, B, and C, are grazing in a field. Deer B is located 61.5 m from d

Question

Three deer, A, B, and C, are grazing in a field. Deer B is located 61.5 m from deer A at an angle of 52.9 ° north of west. Deer C is located 79.3 ° north of east relative to deer A. The distance between deer B and C is 98.2 m. What is the distance between deer A and C?

Explanation / Answer

If you draw out the deer, they are in a triangle. Deer A is a bottom point with deer B upwards and to it's left and deer C upwards and to it's right. You're given the angles from which deer B and deer C are from point A relative to the east/west or horizontal line - if you draw this line, you know that the angles between it must add up to 180. There should be three angles, two known, one unknown. you can use this to figure out angle 'A' --> it's 47.8° Now, you need to use the law of sines: sin(A)/a = sin(B)/b=sin(C)/c we have a, A and c, hence we can solve for angle C --> 27.64 Because we know that the sum of the angles in a triangle must be 180°, we conclude that angle B must be 104.56°. With this information, we have enough to solve for b (aka the distance between deer A and deer C). Using the law of sines, you calculate that b = 128.3. This makes sense, because the angle opposite the longest side must be the greatest.

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