Question

In the given figure, if C is the centre of the circle and $\angle PQC={25}^o$ and $\angle PRC={15}^o$, then $\angle QCR$ is equal to

• A. ${40}^o$
• B. ${60}^o$
• C. ${80}^o$
• D. ${120}^o$

Answer 1

The correct option is C ${80}^o$

Since $CP$, $CQ$ and $CR$ are radii of the same circle then they must be equal, therefore,

$CP=CQ$ and $CP=CR$

Now since $\angle PQC={25}^0$, therefore, $\angle CPQ={25}^0$ and

since $\angle PRC={15}^0$, therefore, $\angle CPR={15}^0$ ........(isoceles triangle property)

Now, $\angle QPR$ can be determined as follows:

$\angle QPR={\angle CPQ+\angle CPR=25}^0+{15}^0={40}^0$

Hence, $\angle QCR={2\angle QPR=2\times 40}^0={80}^0$....(Inscribed angle theorem)