Question

In a cyclic quadrilateral $ABCD,3A=(x+7{)}^o.3B=(y+8{)}^o.3C=(3y+23{)}^o$ and $3D=(4x+12{)}^o$. Find all four angle of the cyclic quadrilateral.

Answer 1

ABCD is cyclic quadrilateral

we know

$\angle A+\angle C=180$ ………….$\left(1\right)$

$\angle B+\angle O=180$ …………..$\left(2\right)$

Equation $\left(1\right)$

$x+7+3y+23=180$

$x+3y=150$ ............$\left(3\right)$

Equation $\left(4\right)$

$y+8+4x+12=180$

$4x+y=160$ …………$\left(4\right)$

Solve $3$ & $4$

$4x+12y=600$

$4x+y=160$

$-$ $-$

_______________

$11y=440$

_______________

$y=40$

$x+40=160$

$4x=120$

$x=30$

So

$\angle A=x+7=30+7=37$

$\angle B=y+8=40+8=48$

$\angle C=3y+23=3\times 40+23=143$

$\angle D=4x+12=4\times 30+12=132$.