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 $\angle D=(4x-7{)}^o$.

Answer 1

We know that the sum of the opposite angles of a cyclic quadrilateral is ${180}^o$. In the cyclic quadrilateral ABCD, angles A and C and the angle B and D form pairs of opposite angles.

$\therefore \angle A+\angle C={180}^o$ and $\angle B+\angle D={180}^o$

$\Rightarrow 2x-1+2y+15=180$

and $y+5+4x-7=180$

$\Rightarrow 2x+2y=166$

and $4x+y=182$

$\Rightarrow x+y=83$ ..(i)
and, $4x+y=182$ ..(ii)

Subtracting equation (i) from equation (ii), we get
$3x=99\Rightarrow x=33$

Substituting $x=33$ in equation (i), we get $y=50$

Hence, $\angle A=(2\times 33-1{)}^o={65}^o,\angle B=(y+5{)}^o=(50+5{)}^o={55}^o$

$\angle C=(2y+15{)}^o=(2\times 50+15{)}^o={115}^o$ and $\angle D=(4\times 33-7{)}^o={125}^o$.