Question

The following figure shows a circle with diameter $AB$ and center at point $O$. If $m$ $\angle$ $CAB=$ ${35}^0$; then $m$ $\angle$ $ABC$ is

• A. ${90}^0$
• B. ${55}^0$
• C. ${60}^0$
• D. ${45}^0$

Answer 1

The correct option is B ${55}^0$

$\angle ACB$ is subtended by the diameter $AB$ to the circumference at $C$.
Therefore, $\angle ACB={90}^o$, since the angle subtended by the diameter of a circle to the circumference $={90}^o$.
Now, by the angle sum property of triangles, we have $\angle ABC+\angle ACB+\angle BAC={180}^o$

$\Rightarrow \angle ABC+{90}^o+{35}^o={180}^o$

$\Rightarrow m\angle ABC={55}^o$