A triangle has sides A, B, and C. Sides A and B are of lengths #4# and #7#, respectively, and the angle between A and B is #(3pi)/8 #. What is the length of side C?

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Answer 1 · 2018-06-24T08:19:05

#c = 6.6# units

#hat C = (3pi)/8, a = 4, b = 7#

Law of cosines #c^2 = a ^2 + b^2 - 2ab cos C#

#:. c^2 = 4^2 + 7^2 - (2 * 4 * 7 * cos ((3pi)/8))#

#c^2 = 43.57#

#c = 6.6 # units

Answer 2 · 2018-06-24T08:22:28

#c=sqrt(65-28*sqrt(2-sqrt(2)))#

Using the Theorem of cosines

#c^2=a^2+b^2-2abcos(gamma)#
and note that

#cos(pi/8)=sqrt(2-sqrt(2))/2#
then we get

#c^2=49+16-2*7*4*cos(3*pi/8)#
so we get

#c^2=65-28sqrt(2-sqrt(2))#