Question

In a cyclic quadrilateral $ABCD$, $\angle A={(2x+11)}^o$, $\angle B={(y+12)}^o$, $\angle C={(3y+6)}^0$ and $\angle D={(5x-25)}^o$, find the angles of the quadrilateral.

• A. $\angle A={75}^o$, $\angle B={45}^o$, $\angle C={105}^o$, $\angle D={135}^o$
• B. $\angle A={50}^o$, $\angle B={50}^o$, $\angle C={150}^o$, $\angle D={110}^o$
• C. $\angle A={80}^o$, $\angle B={30}^o$, $\angle C={120}^o$, $\angle D={130}^o$
• D. $\angle A={60}^o$, $\angle B={40}^o$, $\angle C={130}^o$, $\angle D={130}^o$

Answer 1

The correct option is A $\angle A={75}^o$, $\angle B={45}^o$, $\angle C={105}^o$, $\angle D={135}^o$

In the cyclic quadrilateral sum of opposite angles is 180 degrees
$\angle A+\angle C=180and\angle B+\angle D=180$
According to problem we get equations
$(2x+11)+(3y+6)=\mathrm{180....}\left(1\right)(y+12)+\left(5x25\right)=\mathrm{180......}\left(2\right)2x+3y=\mathrm{163...}\left(1\right)5x+y=\mathrm{193..}\left(2\right)Multiplyeq\left(2\right)by-33(5x+y)=193\Rightarrow -15x-3y=-\mathrm{579...}\left(3\right)Addind\left(1\right)and\left(3\right)weget-13x=-416wegetx=\frac{416}{13}=32andPuttingx=32ineq\left(2\right)wegety=33.Henceanglesare\angle A=2\left(32\right)+11=75\angle B=(y+12)=33+12=45\angle C=(3y+6)=3\left(33\right)+6=105\angle D=(5x-25)=5\left(32\right)-25=135$
$A=75,B=45,C=105o,D=13$