Question

In Fig 10.5 if AOB is a diameter of the circle and AC = BC then $\angle CAB$ is equal to

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

Answer 1

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

We know that the angle at circumference subtended by the diameter of the circle is right angle.

Thus $\angle ACB={90}^0$

In $△ABC$

$AC=BC$ (given)

$=>\angle CAB=\angle CBA$

Now,

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

$=>\angle CAB+\angle CAB=180-90$

$=>\angle CAB=\frac{90}{2}$

$=>\angle CAB={45}^0$