Question

From the diagram, $m\angle PCR=$, and $m\angle PCB=$.

• A. ${130}^{\circ }$
• B. ${50}^{\circ }$
• C. ${120}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

Answer: (A), (B)

${50}^{\circ }$

Here, for the diagram we have to find the value of $m\angle PCR$ and $m\angle PCB$.
Now, $\angle PBC\&\angle QBC$ are lying on the same line.
$\Rightarrow m\angle QBC={180}^{\circ }-{50}^{\circ }={130}^{\circ }$
Now, we know that the sum of all exterior angles $={360}^{\circ }$
$\Rightarrow {100}^{\circ }+{130}^{\circ }+m\angle PCR={360}^{\circ }$
$\Rightarrow m\angle PCR={360}^{\circ }-({100}^{\circ }+{130}^{\circ })$
$\Rightarrow m\angle PCR={360}^{\circ }-{230}^{\circ }={130}^{\circ }$
Also, $\angle PCR$ and $\angle PCB$ are lying on the same line.
$\Rightarrow m\angle PCB={180}^{\circ }-m\angle PCR$
$\Rightarrow m\angle PCB={180}^{\circ }-{130}^{\circ }={50}^{\circ }$
$\therefore m\angle PCB={50}^{\circ }$ and $\angle PCR={130}^{\circ }$.