Question

find the angles of a cyclic quadrilateral ABCD in Angle A = (4x+20 )^0 , Angle B = (3x+5 )^0 , Angle C =(4y)^0 , Angle D = (7y-5)^0​

Answer 1

<A = 120degrees

<B = 80degrees

<C = 60degrees

<D = 100degrees

Explanation:

For a cyclic quadrilateral, the sum of the adjacent angle of the quadrilateral are equal, hence;

<A + <C = 180

<B + <D = 180

Substitute the given values

4x+20+4y = 180

4x+4y =1 60

x + y = 40 ...1

Similarly;

3x+5+7y - 5 = 180

3x+7y = 180 ....2

Solving simultaneously

x + y = 40 ...1 .....* 3

3x+7y = 180 ....2 ...... *1

__________

3x + 3y = 120 ...1 .

3x+7y = 180 ....2

Subtract

3y - 7y = 120 - 180

-4y = -60

y = 15

Since x+y = 40

x + 15 = 40

x = 40 - 15

x = 25

<A = 4x+20

<A = 4(25)+20

<A = 120degrees

<C = 180 - <A

<C = 180 - 120

<C = 60degrees

<B = 3x+5

<B = 3(25)+5

<B = 80degrees

<D = 180 - <B

<D = 180 - 80

<D = 100degrees