Question

In a $△ABC$, if $\angle A={72}^o$ and $\angle B={63}^o$, find the $\angle C$.

Answer 1

We know that, in a $\Delta ABC$,

$A!+B!+C!={180}^{\circ }$

$\Rightarrow {72}^{\circ }+{63}^{\circ }+C!={180}^{\circ }$

$\Rightarrow C!={180}^{\circ }-{72}^{\circ }-{63}^{\circ }$

$={45}^{\circ }$

$\therefore C!={45}^{\circ }$