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
1 answer:
Answer:
<A = 120degrees
<B = 80degrees
<C = 60degrees
<D = 100degrees
Step-by-step 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
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