In a cyclic quadrilateral ABCD, $\angle A = (2x+ 4)^o, \angle B = (y + 3)^o, \angle C = (2y+10)^o$ and $\angle D = (4x - 5)^o$. Find the four angles.

Given:

In a cyclic quadrilateral ABCD, $\angle A = (2x+ 4)^o, \angle B = (y + 3)^o, \angle C = (2y+10)^o$ and $\angle D = (4x - 5)^o$. 

To do:

We have to find 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,

$\angle A+\angle C=180^o$

$ (2x + 4)^o+(2y+10)^o=180^o$

$2x+2y=180^-14^o$

$2(x+y)=166^o$

$x+y=83^o$

$x=83^o-y$.....(i)

$\angle B+\angle D=180^o$

$(y + 3)^o+ (4x - 5)^o=180^o$

$y+4x-2^o=180^o$

$y+4(83^o-y)=180^o+2^o$   (From (i))

$y+332^o-4y=182^o$

$3y=332^o-182^o$

$3y=150^o$

$y=\frac{150^o}{3}$

$y=50^o$

$x=83^o-50^o$   (From (i))

$x=33^o$

This implies,

$\angle A = (2x + 4)^o$

$=2(33^o)+4^o$

$=66^o+4^o$

$=70^o$

$\angle B = (y + 3)^o$

$=50^o+3^o$

$=53^o$

$\angle C = (2y+10)^o$

$=2(50^o)+10^o$

$=100^o+10^o$

$=110^o$

$\angle D = (4x - 5)^o$

$=4(33^o)-5^o$

$=132^o-5^o$

$=127^o$

The four angles are $\angle A=70^o$, $\angle B=53^o$, $\angle C=110^o$ and $\angle D=127^o$. 

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Updated on: 10-Oct-2022

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