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ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
"
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:
$\angle \mathbf{A}=4 y+20^{\circ}$
$=4 \times 25^{\circ}+20^{\circ}$
$=120^{\circ}$
$\angle \mathbf{B}=3 y-5^{\circ}$
$=3 \times 25^{\circ}-5^{\circ}$
$=70^{\circ}$
$\angle \mathbf{C}=-4 x$
$=-4 \times-15^{\circ}$
$=60^{\circ}$
$\angle \mathbf{D}=-7 x+5^{\circ}$
$=-7 \times-15^{\circ}+5^{\circ}$
$=110^{\circ}$