Problem:
In a triangle with sides of lengths a,b and c,(a+b+c)(a+bβc)=3ab. The measure of the angle opposite the side of length c is
Answer Choices:
A. 15β
B. 30β
C. 45β
D. 60β
E. 150β
Solution:
Let ΞΈ be the angle opposite the side of length c. Now
(a+b+c)(a+bβc)(a+b)2βc2a2+b2βabβ=3ab=3ab=c2β
But
a2+b2β2abcosΞΈ=c2
so that ab=2abcosΞΈ,cosΞΈ=21β and ΞΈ=60β.