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Question

In a cyclic quadrilateral ABCD,A=(x+1)° , B=(y+8)°,C=(3y+23)°,,D=(4x+12)° find all the angles

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Solution

The sum of all the angles of the interior angle of a quadrilateral is 360 degrees.
<A = (x+7)° <B = (y+8)° <C=(3y +23)° <D =(4x +12)°

Since we have to solve simultaneously,
We pair up the opposite sides and equal them to 180°.
<A = (x+7)° & <C=(3y +23)°
<B = (y+8)° & <D =(4x +12)°

Now solving simultaneously,
we get,

x+7+3y+23=180°
x+3y+30=180° [180-30]
x+3y=150°·············(i)

y+8+4x+12=180°
y+4x+20=180° [180-20]
y+4x=160°·············(ii)

Now equating the both equations,
x+3y=150°·············(i)
y+4x=160°·············(ii)×3

12x+3y=480·············(ii)
x+3y=150°·············(i)
As solving it the sign changes to minus (-) and 3y gets cancelled,
so,
11x=330
x=
x=30

Now finding the value of y we get,
We take the equation (i) and put the value of x in order to get the value of y,

x+3y=150
30+3y=150
3y=150-30
3y=120
y=
y=40

Now the angles measures,
<A= (x+7)°= (30+7)° =37°
<B= (y+8)° = (40+8)° =48°
<C= (3y+23)° = (3×40+23)° = (120+23)°= 143°
<D=(4x+12)° = (4×30+12)° =(120+12)° =132°

To verify,
Add all the angles to get 360°

like if satisfied

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