ABCD IS A QUADRILATERAL SUCH THAT …

It is given that angles of a cyclic quadrilateral ABCD are given by:
∠A = (4x + 20)°,
∠B = (3x - 5)°,
∠C = (4y)°
and ∠D = (7y + 5)°.
We know that the opposite angles of a cyclic quadrilateral are supplementary.
∠A + ∠C = 180°
4x + 20° + 4y = 180°
4x + 4y – 160° = 0 … (1)
And ∠B + ∠D = 180°
3x – 5 + 7y + 5 = 180°
3x + 7y - 180° = 0… (2)
By elimination method,
Step 1: Multiply equation (1) by 3 and equation (2) by 4 to make the coefficients of x equal.
Then, we get the equations as:
12x + 12y = 480 … (3)
12x + 16y = 540 … (4)
Step 2: Subtract equation (4) from equation (3),
(12x – 12x) + (16y - 12y) = 540 – 480
⇒ 4y = 60
y = 15
Step 3: Substitute value of y in (1),
4x + 4(15) – 160 = 0
⇒ x = 25
Hence, the angles of ABCD are
{tex}\angle{/tex}A = 120°, {tex}\angle{/tex}B = 70°,
{tex}\angle{/tex}C = 60° and {tex}\angle{/tex}D = 110°.