In triangle ABC, a=4,b=6, c=8, how do you find the cosine of each of the angles?

1 Answer
Douglas K.
Jan 10, 2017

Use 3 forms of Law of Cosines :
a^2 = b^2 + c^2 - 2(b)(c)cos(A)
b^2 = a^2 + c^2 - 2(a)(c)cos(B)
c^2 = a^2 + b^2 - 2(a)(b)cos(C)
Solve for cosines

Explanation:

The first form of the Law of Cosines :

a^2 = b^2 + c^2 - 2(b)(c)cos(A)

Solve for cos(A):

cos(A) = (a^2 - b^2 - c^2)/(-2bc)

Substitute, 4 for a, 6 for b, and 8 for c:

cos(A) = (4^2 - 6^2 - 8^2)/(-2(6)(8))

cos(A) = (-84)/-96

cos(A) = 7/8

The second form:

b^2 = a^2 + c^2 - 2(a)(c)cos(B)

Solve for cos(B):

cos(B) = (b^2 - a^2 - c^2)/(-2ac)

Substitute, 4 for a, 6 for b, and 8 for c:

cos(B) = (6^2 - 4^2 - 8^2)/(-2(4)(8))

cos(B) = (-44)/(-64)

cos(B) = 11/16

The third form:

c^2 = a^2 + b^2 - 2(a)(b)cos(C)

Solve for cos(C):

cos(C) = (c^2 - a^2 - b^2)/(-2ab)

Substitute, 4 for a, 6 for b, and 8 for c:

cos(C) = (8^2 - 4^2 - 6^2)/(-2(4)(6))

cos(C) = (12)/(-48)

cos(C) = -1/4

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