The angles of a cyclic quadrilateral \( A B C D \) are
\( \angle \mathrm{A}=(6 x+10)^{\circ}, \angle \mathrm{B}=(5 x)^{\circ}, \angle \mathrm{C}=(x+y)^{\circ}, \angle \mathrm{D}=(3 y-10)^{\circ} \)
Find \( x \) and \( y \), and hence the values of the four angles.
Given:
In a cyclic quadrilateral ABCD, \( \angle \mathrm{A}=(6 x+10)^{\circ}, \angle \mathrm{B}=(5 x)^{\circ}, \angle \mathrm{C}=(x+y)^{\circ}, \angle \mathrm{D}=(3 y-10)^{\circ} \).
To do:
We have to find \( x \) and \( y \), and hence the values of the four angles.
Solution:
We know that,
Sum of the angles in a quadrilateral is $360^o$.
Sum of the opposite angles in a cyclic quadrilateral is $180^o$.
Therefore,
$\angle A+\angle C=180^o$
$ (6x +10)^o+(x+y)^o=180^o$
$7x+y=180^o-10^o$
$7x+y=170^o$
$y=170^o-7x$.......(i)
$\angle B+\angle D=180^o$
$(5x)^o+ (3y- 10)^o=180^o$
$5x+3y-10^o=180^o$
$5x+3y=180^o+10^o$
$5x+3(170^o-7x)=190^o$ (From (i))
$5x+510^o-21x=190^o$
$16x=510^o-190^o$
$16x=320^o$
$x=\frac{320^o}{16}$
$x=20^o$
$y=170^o-7(20^o)$ (From (i))
$y=170^o-140^o$
$y=30^o$
This implies,
$\angle A = (6x + 10)^o$
$=6(20^o)+10^o$
$=120^o+10^o$
$=130^o$
$\angle B = 5x^o$
$=5(20^o)$
$=100^o$
$\angle C = (x+y)^o$
$=20^o+30^o$
$=50^o$
$\angle D = (3y - 10)^o$
$=3(30^o)-10^o$
$=90^o-10^o$
$=80^o$
The four angles are $\angle A=130^o$, $\angle B=100^o$, $\angle C=50^o$ and $\angle D=80^o$.
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