Find the four angles of a cyclic quadrilateral `A B C D` in which `/_A=(2x-1)o` , `/_B=(y+5)o\ \ /_C=(2y+15)o` and `/_D=(4x-7)o` .

Options

A. ∠
A
=
75
∘
,
∠
B
=
55
∘
,
∠
C
=
115
∘
and
∠
D
=
115
∘
B. ∠
A
=
65
∘
,
∠
B
=
55
∘
,
∠
C
=
125
∘
and
∠
D
=
115
∘
C. ∠
A
=
65
∘
,
∠
B
=
55
∘
,
∠
C
=
115
∘
and
∠
D
=
115
∘
D. ∠
A
=
65
∘
,
∠
B
=
55
∘
,
∠
C
=
115
∘
and
∠
D
=
125
∘

Answer

D

Explanation

We know that the sum of the opposite angles of a cyclic quadrilateral is ` 180^(@)`.
` therefore angle A + angle C = 180^(@) and angle B + angle D = 180^(@)`
Now, ` angle A + angle C = 180^(@) rArr ( 2x - 1 ) + ( 2y + 15 ) = 180`
` " " rArr 2 (x + y ) = 166`
` " " rArr x+y = 83" " ` ... (i)
And, ` angle B + angle D = 180^(@) rArr ( y + 5) + ( 4 x - 7) = 180 `
`" " rArr 4 x + y = 182 " " `... (ii)
On subtracting (i) from (ii), we get
` 3 x = 182 - 83 = 99 rArr x = 33 `.
Putting ` x = 33 ` in (i), we get
` 33 + y = 83 rArr y = ( 83 - 33) rArr y = 50 `
Thus, ` x = 33 and y = 50 `
` therefore angle A = { ( 2 xx 33) - 1}^(@) = 65 ^(@), angle B = ( 50 + 5)^(@) = 55 ^(@)`
` angle C = {( 2 xx 50 ) + 15}^(@) = 115^(@) and angle D = {( 4 xx 33 ) - 7}^(@) = 125 ^(@) `
Hence, ` angle A = 65 ^(@), angle B = 55^(@), angle C = 115^(@) and angle D = 115^(@)`.