Question

Two chords AB and CD of a circle cut each other when produced outside the circle at P. AD and BC are joined. If $\angle PAD={30}^o$ and $\angle CPA={45}^o$. Find $\angle CBP.$

• A. ${105}^o$
• B. ${115}^o$
• C. ${135}^o$
• D. None of these

Answer 1

The correct option is A ${105}^o$

$\angle PAD=\angle PCB={30}^{\circ }$
In $△CBP$
$\angle C+\angle P+\angle B={180}^{\circ }$
$={30}^{\circ }+{45}^{\circ }+\angle B={180}^{\circ }$
$\angle B={180}^{\circ }-{75}^{\circ }$
$\angle B={105}^{\circ }$
$\angle CBP={105}^{\circ }$