How do you find the exact value of the third side given #triangle ABC#, a=6, b=4, #mangleC=(2pi)/3#?

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Answer 1 · 2017-11-19T17:28:02

#c= 2sqrt(19)#

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Labelling triangle as diagram:

Using the Cosine Rule:

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

#c^2=(6)^2+(4)^2-(2)(6)(4)*cos((2pi)/3)#

#c^2=(6)^2+(4)^2-(2)(6)(4)*(-1/2)#

#c^2=52-(-24)=76#

#c^2=76=>c=sqrt(76)=2sqrt(19)#