Question

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

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

Answer 1

The correct option is C ${48}^{\circ }$

As the $\angle BDC$ and $\angle ACB$ are in the same segment, thus they are equal.

But $\angle BDC={42}^{\circ }$

therefore $\angle BAC={42}^{\circ }$
In $△BAC$

$\angle BAC+\angle BCA={90}^{\circ }$

$\angle BCA=90-42=48$

$\therefore \angle ACB={48}^{\circ }$