Question

In the given circle ABCD, O is the centre and $\angle BDC$ = ${42}^0$, The $\angle ACB$ is equal to

• A. ${42}^0$
• B. ${45}^0$
• C. ${48}^0$
• D. ${60}^0$

Answer 1

Answer: (C)

${48}^0$

In the given figure,

$\angle BDC={42}^o$

$\angle BAC={42}^o$ (Angles subtended by the same arc in same segment of a circle are equal )

Now, since $\angle ABC={90}^o$ (angle in semicircle is a right angle)

By angle sum property,

$\angle BAC+\angle ABC+\angle ACB={180}^o$

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

$\Rightarrow \angle ACB={48}^o$