Question

Find all the angles of the given cyclic quadrilateral $ABCD$ in the figure.

Answer 1

In a cyclic quadrilateral $ABCD$.
$\angle B+\angle D={180}^{\circ }\dots ($ Sum of the opposite angles of a cyclic quadrilateral is ${180}^{\circ }$ )
$6x-4^{\circ }+7x+2^{\circ }={180}^{\circ }$
$13x-2^{\circ }={180}^{\circ }$
$13x={182}^{\circ }$
$x={182}^{\circ }$
$x=\frac{{182}^{\circ }}{13}$
$x={14}^{\circ }$
$\angle B=6x-4^{\circ }$
$\begin{array}{cc}& =6\left(14\right)-4^{\circ }\\ & =84-4\\ & ={80}^{\circ }\end{array}$
$\begin{array}{cc}& \angle D=7x+2^{\circ }\\ & =7\left(14\right)+2^{\circ }\\ & =98+2\\ & ={100}^{\circ }\end{array}$
$2y+4^{\circ }+4y-4^{\circ }={180}^{\circ }\dots ($ Sum of the opposite angles of a cyclic quadrilateral is ${180}^{\circ }$ )
$$\begin{array}{cc}& 6y={180}^{\circ }\\ & y=\frac{{180}^{\circ }}{6}\\ & ={30}^{\circ }\\ & \angle A=2y+4^{\circ }\\ & =2\left(30\right)+4^{\circ }\\ & ={64}^{\circ }\\ & \angle C=4y-4^{\circ }\\ & =4\left(30\right)-4^{\circ }\\ & ={120}^{\circ }-4^{\circ }\\ & ={116}^{\circ }\\ & \therefore \angle A={64}^{\circ },\angle B={80}^{\circ },\angle C={116}^{\circ },\angle D={100}^{\circ }.\end{array}$$