Question

In the given figure if O is the centre of the circle, then $\angle CAB$ equals

• A. ${90}^{\circ }$
• B. ${75}^{\circ }$
• C. ${50}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

Answer: (D)

${45}^{\circ }$

Given, O is the centre of the circle.

Hence AB is the diameter of the given circle.

Since angle in the semicircle is 90 degrees, $\angle ACB={90}^{\circ }$

Also, given, AC = BC.

We know that, in a triangle angles opposite to equal sides are equal.

Hence $\angle CAB=\angle CBA=x^{\circ }$ (say)

Now, by applying angle sum property to triangle ABC, we get

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

i.e., ${90}^{\circ }+x^{\circ }+x^{\circ }={180}^{\circ }$

$⟹x^{\circ }={45}^{\circ }$

Therefore $\angle CAB={45}^{\circ }$