Question

In a cyclic quadrilateral ABCD,$\begin{array}{c},\angle A={\left(2x+4\right)}^o,\angle B={\left(y+3\right)}^o,\angle C={\left(2y+10\right)}^o\end{array}$and$\angle D={\left(4x-5\right)}^o.$ Find the smallest angle.

Answer 1

$\text{In cyclic quadrilateral ABCD,}$

$\text{Sum of opposite angles is}{180}^{\circ }$

$\angle A+\angle C={180}^{\circ }$

$2x+4+2y+10={180}^{\circ }-\left(1\right)$

$\angle B+\angle D={180}^{\circ }$

$y+3+4x-5={180}^{\circ }-\left(2\right)$

$\text{Multiply eqn (1) by 2 and sub}\text{. (2) from (1),}$

$\left(4x+4y+28={360}^{\circ }\right)-\left(4x+y-2={180}^{\circ }\right)$

$3y+30={180}^{\circ }$

$y=\frac{150}{3}={50}^{\circ }$

$\left(4x+50-2\right)={180}^{\circ }$

$x=\frac{180-48}{4}={33}^{\circ }$

$\therefore \angle A={70}^{\circ }\angle B={53}^{\circ }$

$\angle C={110}^{\circ }\angle D={180}^{\circ }$