Question

Three deer, A, B, and C, are grazing in a field. Deer B is located 62m from deer A at an angle of 51° north of West. Deer C is located 77° north of east relative to deer A. The distance between deer B and C is 95m. What is the distance between deer A and C?

Answer 1

Set deer A's position to be the origin. Let $c$ be the distance from deer A to deer C. We're given that deer B is 95 m away from deer C, which means the length of the vector $B-C$ is 95 (or $C-B$). Then

$|B-C|^2=(B-C)\cdot(B-C)=B\cdot B-2B\cdot C+C\cdot C=|B|^2-2B\cdot C+|C|^2$

$|B-C|^2=|B|^2-2|B||C|\cos(180-77-51)^\circ+c^2$

$95^2=62^2-2(62)(c)\cos52^\circ+c^2$

$c^2-124\cos52^\circ c-5181=0\implies c=120\,\mathrm m$

Answer 2

Based on the data given, the distance between deer A and C is 119.6 m

What is the distance between A and C?

Using the information provided, a triangle ABC is formed

• Angle at A = 180° - 128° = 52°
• length of side AB, given as c = 62 m
• length of side BC, given as a = 95 m
• length of side AC, given as b = ?

Using sine rule to find the angle at B

• a/sinA = c/sinC

C = sin¹ (c * Sin A/a)

C = sin¹ (62 * sin 52/95)

C = 31°

Therefore;

angle at B = 180 - (52 + 31)

angle at B = 97°

Using cosine rule to find side AC

• b² = a² + c² - 2ac * cosB

b² = 95² + 62² - 2 * 95 * 62 * cos97

b = 119.6 m

Therefore, the distance between deer A and C is 119.6 m

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