ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
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Given:

ABCD is a cyclic quadrilateral.

To do:

We have to find the angles of the cyclic quadrilateral.

Solution:

We know that,

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

This implies,

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

$3 y-5^{\circ}+(-7 x+5^{\circ}) =180^{\circ}$

$3 y-5^{\circ}-7 x+5^{\circ} =180^{\circ}$

$-7 x+3 y =180^{\circ}$

$7 x-3 y =-180^{\circ}$...........(i)

Similarly,

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

$4 y+20^{\circ}+(-4x) =180^{\circ}$

$4y-4x=180^{\circ}-20^{\circ}$

$4 x-4 y =-160^{\circ}$

$4(x-y) =4(-40^{\circ})$

$x-y=-40^{\circ}$........(ii)

Multiplying (ii) by 3 and subtracting the result from (i), we get,

$7x-4 x=-180^{\circ}+120^{\circ}$

$4 x=-60^{\circ}$

$x=-\frac{60^{\circ}}{4}$

$x=-15^{\circ}$

Substituting of $x=-15^{\circ}$ in (ii), we get,

$-15^{\circ}-y=-40^{\circ}$

$y=(40-15)^{\circ}$

$y=25^{\circ}$

Hence, the angles of the cyclic quadrilateral are:

$\angle \mathbf{A}=4 y+20^{\circ}$

$=4 \times 25^{\circ}+20^{\circ}$

$=120^{\circ}$

$\angle \mathbf{B}=3 y-5^{\circ}$

$=3 \times 25^{\circ}-5^{\circ}$

$=70^{\circ}$

$\angle \mathbf{C}=-4 x$

$=-4 \times-15^{\circ}$

$=60^{\circ}$

$\angle \mathbf{D}=-7 x+5^{\circ}$

$=-7 \times-15^{\circ}+5^{\circ}$

$=110^{\circ}$