Question

In the given figure AOB is a diameter of the circle and AC=BC The $\angle CAB$ is

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${45}^{\circ }$

$\Rightarrow$ In $△ABC$,

$\Rightarrow$ $\angle BCA={90}^o$ [ Angle inscribed in semi-circle ]

$\Rightarrow$ $AC=BC$ [ Given ]

$\therefore$ $\angle ABC=\angle CAB$ [ Opposite angles of equal sides are equal ]

$\therefore$ $\angle CAB+\angle ABC+\angle BCA={180}^o$

$\therefore$ $\angle CAB+\angle CAB={90}^o={180}^o$

$\therefore$ $2\angle CAB={180}^o-{90}^o$

$\therefore$ $2\angle CAB={90}^o$

$\therefore$ $\angle CAB={45}^o$