Question

In the figure given above, if $\angle BAD={60}^{\circ },\angle ADC={105}^{\circ }$, then what is $\angle DPC$ equal to?

• A. ${40}^{\circ }$
• B. ${45}^{\circ }$
• C. ${50}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

Answer: (B)

${45}^{\circ }$

Given, $\angle BAD={60}^0,\angle ADC={105}^0$

We need to find measure of $\angle DPC$

$\angle BAD=\angle PAD={60}^{\circ }$
$\angle PDC={180}^{\circ }-{105}^{\circ }={75}^{\circ }$
$\therefore \angle DPC={180}^{\circ }-({60}^{\circ }+{75}^{\circ })$
$={45}^{\circ }$