Question

in a cyclic quadrilateral ABCD.Angle A=2x+4 ,angle B =y+3 ,angle C=2y+10 and angle D=4x+5.Find x and y.

Answer 1

In a cyclic quadrilateral the sum of opposite angles is 180
so <A+<C=180°
and <B + <D = 180°
so we get
2x+4+2y+10=180
=>2x + 2y +14=180
=>x+y=83 -–––––-【1】
& y +3+4x+5=180
=>4x+y+8=180
=>4x+y=172––––––-【2】

Solve the equation to get the answer.

hope it helps....

Answer 2

Given:

A cyclic quadrilateral ABCD has

$\angle A = 2x + 4$

$\angle B = y + 3$

$\angle C = 2y + 10$

$\angle D = 4x + 5$

To find:

The values of “x” and “y”

Answer:

In cyclic quadrilateral opposite angles sum is equal to $180 ^{\circ}$

$\angle A + \angle C = 180 ^{\circ} \ and \ \angle B + \angle D = 180  ^{\circ}$

Now,

$2x + 4 +2y +10 = 180 ^{\circ}$

$2x + 2y = 180 ^{\circ} -14^{\circ}$

$2x + 2y = 166^{\circ}$

$x + y = 83 ^{\circ}....... (1)$

Similarly,

$y + 3 + 4x - 5 = 180 ^{\circ}$

$4x + y = 182 ^{\circ} ....(2)$

By solving two equations, we will get x and y,

$x + y - 83 ^{\circ} = 4x + y - 182 ^{\circ}$

$3x = 99 ^{\circ}$

$x = 33 ^{\circ}$

Substitute x value in equation

$33 ^{\circ} + y = 83$

$y = 50 ^{\circ}$

Now calculate,

$\angle A = 2 \times 33 + 4 = 70 ^{\circ}$

$\angle B = 50 + 3 = 53 ^{\circ}$

$\angle C = 2 \times 50 + 10 = 110 ^{\circ}$

$\angle D = 4 \times 33 + 5 = 137 ^{\circ}$