# 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.

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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 $\angle D=80^o$.

Updated on 10-Oct-2022 13:27:21