Question

ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.

• A. ${120}^{\circ },{70}^{\circ },{60}^{\circ },{110}^{\circ }$
• B. ${130}^{\circ },{75}^{\circ },{50}^{\circ },{105}^{\circ }$
• C. ${100}^{\circ },{80}^{\circ },{60}^{\circ },{120}^{\circ }$
• D. None of the above

Answer 1

The correct option is A

${120}^{\circ },{70}^{\circ },{60}^{\circ },{110}^{\circ }$

We know that the sum of the measure of opposite angles in a cyclic quadrilateral ${180}^{\circ }$ Therefore, $\Rightarrow \angle A+\angle C={180}^{\circ }$.
$\Rightarrow 4y+20-4x={180}^{\circ }$
$\Rightarrow -4x+4y={160}^{\circ }$
$\Rightarrow x-y=-{40}^{\circ }$ ....(i)

Also, $\angle B+\angle D={180}^{\circ }$
$\Rightarrow 3y-5-7x+5={180}^{\circ }$
$\Rightarrow -7x+3y={180}^{\circ }$ .....(ii)

Multiplying equation (i) by 3, we obtain
$3x-3y=-{120}^{\circ }$ ......(iii)
Adding equations (ii) and (iii) we obtain,
$\Rightarrow -7x+3x=(180-120{)}^{\circ }$
$\Rightarrow -4x={60}^{\circ }$
$\Rightarrow x=-{15}^{\circ }$

By using equation (i), we obtain,
$x-y=-{40}^{\circ }$
$\Rightarrow -15-y=-{40}^{\circ }$
$\Rightarrow y=-15+40={25}^{\circ }$

$\angle A=4y+20=4\left(25\right)+20={120}^{\circ }$
$\angle B=3y-5=3\left(25\right)-5={70}^{\circ }$
$\angle C=-4x=-4(-15)={60}^{\circ }$
$\angle D=-7x+5=-7(-15)+5={110}^{\circ }$