Problem:
In a triangle with sides of lengths a,b and c,(a+b+c)(a+b−c)=3ab. The measure of the angle opposite the side of length c is
Answer Choices:
A. 15∘
B. 30∘
C. 45∘
D. 60∘
E. 150∘
Solution:
Let θ be the angle opposite the side of length c. Now
(a+b+c)(a+b−c)(a+b)2−c2a2+b2−ab=3ab=3ab=c2
But
a2+b2−2abcosθ=c2
so that ab=2abcosθ,cosθ=21 and θ=60∘.