ABCD is a cyclic quadrilateral such that $\angle A = (4y + 20)^o, \angle B = (3y – 5)^o, \angle C = (4x)^o$ and $\angle D = (7x + 5)^o$. Find the four angles.

Given:

ABCD is a cyclic quadrilateral such that $\angle A = (4y + 20)^o, \angle B = (3y – 5)^o, \angle C = (4x)^o$ and $\angle D = (7x + 5)^o$.

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,

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

$ (4y + 20)^o+(4x)^o=180^o$

$4y+4x=180^-20^o$

$4(x+y)=160^o$

$x+y=40^o$

$x=40^o-y$.....(i)

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

$(3y – 5)^o+ (7x + 5)^o=180^o$

$3y+7x=180^o$

$3y+7(40^o-y)=180^o$   (From (i))

$3y+280^o-7y=180^o$

$4y=280^o-180^o$

$4y=100^o$

$y=\frac{100^o}{4}$

$y=25^o$

$x=40^o-25^o$   (From (i))

$x=15^o$

This implies,

$\angle A = (4y + 20)^o$

$=4(25^o)+20^o$

$=100^o+20^o$

$=120^o$

$\angle B = (3y – 5)^o$

$=3(25^o)-5^o$

$=75^o-5^o$

$=70^o$

$\angle C = (4x)^o$

$=4(15^o)$

$=60^o$

$\angle D = (7x + 5)^o$

$=7(15^o)+5^o$

$=105^o+5^o$

$=110^o$

The four angles are $\angle A=120^o$, $\angle B=70^o$, $\angle C=60^o$ and $\angle D=110^o$.

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Updated on: 10-Oct-2022

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