Question

The sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at P, the sides AD and BC are produced to meet at Q. If $\angle ADC={85}^{\circ }$ and $\angle BPC={40}^{\circ }$, then $\angle CQD$ equals

• A. 30${}^{\circ }$
• B. 45${}^{\circ }$
• C. 60${}^{\circ }$
• D. 75${}^{\circ }$

Answer 1

The correct option is A 30${}^{\circ }$

$\angle ADC+\angle ABC=180$

$\to$ supplementary due to being cyclically opposite angles

so $\angle ABC={95}^o$

and $\angle ABC+\angle CBP=180$

$\to$ supplementary

$\angle CBP={85}^o$

then in triangle $\Delta BPC$

$\angle B+\angle P+\angle C={180}^o$

$\angle C={55}^o$

$\angle ADC+\angle QDC=180$

$\to$ supplementary

$\angle QDC=95$

In triangle $\Delta DQC$

$\Rightarrow \angle Q+\angle D+\angle C={180}^o$

so $\angle Q={30}^o$.