Question

Find the four angles of a cyclic quadrilateral ABCD in which $\angle A=(2x-1{)}^o,\angle B=(y+5{)}^o,\angle C=(2y+15{)}^o$ and $\angle D(4x-7{)}^o$

• A. $\angle A={25}^o,\angle B={45}^o,\angle C={105}^o,\angle D={135}^o$
• B. $\angle A={35}^o,\angle B={75}^o,\angle C={95}^o,\angle D={135}^o$
• C. $\angle A={45}^o,\angle B={75}^o,\angle C={95}^o,\angle D={125}^o$
• D. $\angle A={65}^o,\angle B={55}^o,\angle C={115}^o,\angle D={125}^o$

Answer 1

Answer: (D)

$\angle A={65}^o,\angle B={55}^o,\angle C={115}^o,\angle D={125}^o$

Given, $\angle A={(2x-1)}^{\circ }$, $\angle B={(y+5)}^{\circ }$, $\angle C={(2y+15)}^{\circ }$ and $\angle D={(4x-7)}^{\circ }$.

In cyclic quadrilateral, the sum of opposite angles $={180}^o$
$\Rightarrow \angle A+\angle C={180}^o$
$\Rightarrow (2x-1)+(2y+15)={180}^{\circ }$
$\Rightarrow 2x+2y={166}^{\circ }\Rightarrow x+y=83$ .....(1)

Also, $\angle B+\angle D={180}^o$
$\Rightarrow (y+5)+(4x-7)={180}^{\circ }$
$\to 4x+y={182}^{\circ }$ ....(2)

Substract (2) from (1), we get,
$\Rightarrow -3x=-{99}^o\Rightarrow x={33}^o$.

Now, substitute $x=33$ in (1), we get,
$\Rightarrow {33}^o+y={83}^o\Rightarrow y={50}^o$.

Hence,
$\angle A=(2x-1)=2\times 33-1={65}^{\circ }$
$\angle B=(y+5)=50+5={55}^{\circ }$
$\angle C=(2y+15)=2\times 50+15={115}^{\circ }$
$\angle D=(4x-7)=4\times 33-7={125}^{\circ }$.

Therefore, option $D$ is correct.