Find all the interior angles of the cyclic quadrilateral ABCD.

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Solution

We know that the sum of the measure of opposite angles in a cyclic quadrilateral $180^{\circ}$ Therefore, $\Rightarrow ∠ A + ∠ C = 180^{\circ}$ .
$\Rightarrow 4 y + 20 - 4 x = 180^{\circ}$
$\Rightarrow - 4 x + 4 y = 160^{\circ}$
$\Rightarrow x - y = - 40^{\circ}$ ....(i)

Also, $∠ B + ∠ D = 180^{\circ}$
$\Rightarrow 3 y - 5 - 7 x + 5 = 180^{\circ}$
$\Rightarrow - 7 x + 3 y = 180^{\circ}$ .....(ii)

Multiplying equation (i) by 3, we obtain
$3 x - 3 y = - 120^{\circ}$ ......(iii)
Adding equations (ii) and (iii) we obtain,
$\Rightarrow - 7 x + 3 x = (180 - 120)^{\circ}$
$\Rightarrow - 4 x = 60^{\circ}$
$\Rightarrow x = - 15^{\circ}$

By using equation (i), we obtain,
$x - y = - 40^{\circ}$
$\Rightarrow - 15 - y = - 40^{\circ}$
$\Rightarrow y = - 15 + 40 = 25^{\circ}$

$∠ A = 4 y + 20 = 4 (25) + 20 = 120^{\circ}$
$∠ B = 3 y - 5 = 3 (25) - 5 = 70^{\circ}$
$∠ C = - 4 x = - 4 (- 15) = 60^{\circ}$
$∠ D = - 7 x + 5 = - 7 (- 15) + 5 = 110^{\circ}$

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