Question

"In the given figure ,$\angle BAC$ is ${62}^{\circ }$ says Sita . Do you agree with her?

Answer 1

Consider the given figure,

$\angle ABC={63}^0$

Let $\angle ABC=x^0$ and $\angle ACB=y^0$

Now,

$\angle ACB+{125}^0={180}^0$

$\angle ACB={180}^0-{125}^0$

$\angle ACB={55}^0$

In $\Delta ABC$

$\angle ABC+\angle ACB+\angle CAB={180}^0$

${63}^0+{55}^0+\angle CAB={180}^0$

$\angle CAB={180}^0-{118}^0$

$\angle CAB={62}^0$