Question

The angles of a cyclic quadrilateral $ABCD$ are
$\angle A=(6x+10{)}^{\circ },\angle B=(5x{)}^{\circ }$
$\angle C=(x+y{)}^{\circ },\angle D=(3y-10{)}^{\circ }$
Find $x$ and $y,$ and hence the values of the four angles.

Answer 1

Forming an equation by using the properties of cyclic quadrilateral.​
Given, $ABCD$ is a cyclic Quadrilateral.
Since, the sum of opposite angle of a cyclic quadrilateral measures ${180}^{\circ }.$
Therefore,
$\angle A+\angle C={180}^{\circ }\text{and}\angle B+\angle D={180}^{\circ }$
$\Rightarrow (6x+10)+(x+y)={180}^{\circ }$
$7x+y={180}^{\circ }-{10}^{\circ }$
$\Rightarrow 7x+y={170}^{\circ }......\left(1\right)$
and
$5x+(3y-10)={180}^{\circ }$
$5x+3y={180}^{\circ }+10$
$5x+3y={190}^{\circ }.....\left(2\right)$

Finding the values of angles of quadrilateral.
From equation(1), we get
$y={170}^{\circ }-7x.....\left(3\right)$
Substituting the value from (3) to (2)
$5x+3({170}^{\circ }-7x)={190}^{\circ }$
$\Rightarrow 5x+{510}^{\circ }-21x={190}^{\circ }$
$\Rightarrow -16x={190}^{\circ }-{510}^{\circ }$
$\Rightarrow -16x=-{320}^{\circ }$
$\Rightarrow x=\frac{-320}{-16}$
$\Rightarrow x={20}^{\circ }$
Substituting the value of $x$ in equation (3) we get,
$y={170}^{\circ }-7\left({20}^{\circ }\right)$
$y={170}^{\circ }-{140}^{\circ }$
$y={30}^{\circ }$
Therefore, values of $x$ and $y$ are ${20}^{\circ }$ are ${30}^{\circ }$ respectively.
Now,
$\angle A=6\left({20}^{\circ }\right)+{10}^{\circ }={120}^{\circ }+{10}^{\circ }={130}^{\circ }$
$\angle B=5\left({20}^{\circ }\right)={100}^{\circ }$
$\angle C={20}^{\circ }+{30}^{\circ }={50}^{\circ }$
$\angle D=3\left({30}^{\circ }\right)-{10}^{\circ }=(90-{10}^{\circ })={80}^{\circ }$

Hence the values of $x$ and $y$ are ${20}^{\circ }$ and ${30}^{\circ }$ respectively and the values of the four angles are ${130}^{\circ },{100}^{\circ },{50}^{\circ }\text{and}{80}^{\circ }.$

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