Find the four angles of cyclic quadrilateral ABCD in which /_A= (2x +4)°, /_B = ( y + 3 )°, /_C = ( 2y + 10 )° and /_D = ( 4x - 5 )°.
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Sum of all the angles of a quadrilateral is 360
So, 2x +4+ y+3+2y+10+4x-5=360
=> 6x + 3y + 12 = 360
=>3(2x +y) = 360-12
=> 2x + y = 348/3
=> 2x + y = 116 ---------(i)
And,
(2x+4) + (2y+10)=180 [sum of opposite angles of cyclic quad. is 180]
=> (2x + y) +y + 4+10=180
=> 116 + y + 14 = 180
=> y = 180-130
=> y = 50° ----------(ii)
Using eqn(ii) in eqn (i)
2x + 50 = 116
=> 2x = 116-50
=> x = 66/2= 33
=> x = 33°
Now,
A = 2×33+4=70°
B = 50+3=53°
C = 2×50+10= 110°
D = 4×33-5= 127°
So, 2x +4+ y+3+2y+10+4x-5=360
=> 6x + 3y + 12 = 360
=>3(2x +y) = 360-12
=> 2x + y = 348/3
=> 2x + y = 116 ---------(i)
And,
(2x+4) + (2y+10)=180 [sum of opposite angles of cyclic quad. is 180]
=> (2x + y) +y + 4+10=180
=> 116 + y + 14 = 180
=> y = 180-130
=> y = 50° ----------(ii)
Using eqn(ii) in eqn (i)
2x + 50 = 116
=> 2x = 116-50
=> x = 66/2= 33
=> x = 33°
Now,
A = 2×33+4=70°
B = 50+3=53°
C = 2×50+10= 110°
D = 4×33-5= 127°
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