Question

The point $A,B,C$ and $D$ are lying on a circle with centre $O$. Find the measure of $\angle ACB$.

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

Answer 1

The correct option is C ${48}^0$

$OA=OB=OC=OD=radius$

In $△OCD$

$OC=OD$

$\angle OCD=\angle ODC={42}^{\circ }$

$\angle DOC=180–42–42={96}^{\circ }$

$\angle DOC+\angle COB=180$

$\angle COB=180–96={84}^{\circ }$

In $△COB$

$OC=OB$

$\angle OCB=\angle OBC$

$\angle OCB+\angle COB+\angle OBC=180$

$2\times \angle OCB=180–84$

$\angle OCB=48=\angle ACB$