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

1 Answer
Jan 10, 2017

Answer:

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#