Question

In fig. if $\angle ACB={40}^{\circ }$, $\angle DPB={120}^{\circ }$, then $y$ is equal to:

• A. ${10}^{\circ }$
• B. ${20}^{\circ }$
• C. ${15}^{\circ }$
• D. ${25}^{\circ }$

Answer 1

The correct option is B ${20}^{\circ }$

$\angle D=\angle C$ [$\because \angle s$ in the same segment of a circle are equal]
But $\angle C={40}^{\circ }$ (Given)
$\therefore \angle D={40}^{\circ }$
In $\Delta BPD$, we have
$\angle BPD+\angle D+\angle B={180}^{\circ }$ [$\because$ sum of three $\angle s$ of a $\Delta ={180}^{\circ }$]
$\Rightarrow {120}^{\circ }+{40}^{\circ }+y={180}^{\circ }$
$\Rightarrow y={180}^{\circ }-{160}^{\circ }={20}^{\circ }$