Question

Suppose an isosceles triangle abc has a=pi/4 and b=c=4. What is the length of a^2?

Answer 1

The law of cosines states that, if $\alpha$ is the angle between sides b and c,

$a^2 = b^2+c^2-2bc\cos(\alpha)$

So, plugging our values, we have

$a^2 = 16+16-2\cdot 4\cdot 4\cdot\cos\left(\dfrac{\pi}{4}\right)$

This is equal to

$a^2 = 32-32\cdot\dfrac{\sqrt{2}}{2} = 32-16\sqrt{2}$