Question

In the figure, O is the centre of the circle and $\angle BDC={42}^o$. The measure of $\angle ACB$ is

• A. ${42}^o$
• B. ${48}^o$
• C. ${58}^o$
• D. ${52}^o$

Answer 1

The correct option is B

${48}^o$

In the figure, O is the centre of the circle and $\angle BDC={42}^o$.

$\angle ABC={90}^o$ (Angle in a semicircle)

Also, $\angle BACand\angle BDC$ are in the same segment of the circle.

$\therefore \angle BAC=\angle BDC={42}^o$

Now in $\Delta ABC$, $\angle A+\angle ABC+\angle ACB={180}^o$ (Sum of angles of a triangle)

$\Rightarrow {42}^o+{90}^o+\angle ACB={180}^o$

$\Rightarrow {132}^o+\angle ACB={180}^o$

$\Rightarrow \angle ACB={180}^o-{132}^o={48}^o$