#### Answer

$c^{2}= a^{2} +b^{2}- 2\times a \times b \times 0= c^{2}= a^{2} +b^{2}$

#### Work Step by Step

If $\gamma = 90^{\circ}$, then
$\cos \gamma =0$
Therefore, if we replace $\cos \gamma$ by 0 in the equation, we get
$c^{2}= a^{2} +b^{2}- 2\times a \times b \times 0$
$ c^{2}= a^{2} +b^{2}$
And that is the Pythagorean Theorem