Question

ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.

Answer 1

The sum of the opposite angles in cyclic quadrilateral is ${180}^{\circ }$

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

$\Rightarrow 4y+20-4x=180$

$\Rightarrow 4y-4x=180-20$

$\Rightarrow 4y-4x=160$

$\Rightarrow x-y=-40$....... (i)

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

$\Rightarrow 3y-5-7x+5=180$

$\Rightarrow -7x+3y=180$....... (ii)

Multiplying equation (i) by 3, we obtain

$\Rightarrow 3x-3y=-120$........ (iii)

Adding equations (ii) and (iii), we obtain

$\Rightarrow -7x+3x=180-120$

$\Rightarrow -4x=60$

$\Rightarrow x=-15$

By using equation (i), we obtain

$\Rightarrow x-y=-40$

$\Rightarrow -15-y=-40$

$\Rightarrow y=-15+40=25$

$\angle A=4y+20=4\times 25+20={120}^{\circ }$

$\angle B=3y-5=3\times 25-5={70}^{\circ }$

$\angle C=-4x=-4\times -15={60}^{\circ }$

$\angle D=-7x+5=-7\times -15+5={110}^{\circ }$