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 the $\angle ADC={85}^{\circ }$ and $\angle BPC={40}^{\circ }$, then the $\angle CQD$ equals.

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

Answer 1

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

ABCD is a cyclic quadrilateral, therefore

$\angle D+\angle B=180°\Rightarrow 85°+\angle B=180°\Rightarrow \angle B=180°-85°=95°$

$\angle CBA+\angle CBP=180°$ [Linear pair]

$95°+\angle CBP=180°\Rightarrow \angle CBP=85°$

In $△BPC$, sum of all angles $=180°$

$\angle CBP+\angle BPC+\angle PCB=180°\Rightarrow 85°+40°+\angle PCB=180°$

$\angle PCB=180°-125°=55°$

$\angle PCB+\angle DCB=180°$ [linear pair] $\Rightarrow 55°+\angle DCB=180°\Rightarrow \angle DCB=125°$

$\angle QDC+\angle ADC=180°$ [linear pair]$\angle QDC+85°=180°\Rightarrow \angle QDC=95°$

In $△QDC$, sum of all angles$=180°$

$\Rightarrow \angle QDC+\angle QCD+\angle CQD=180°\Rightarrow 95°+(180°-125°)+\angle CQD=180°$

$\Rightarrow 95°+55°+\angle CQD=180°\Rightarrow \angle CQD=180°-150°=30°$