# Find the angles of a cyclic quadrilateral ABCD in which∠A = (4x + 20)°, ∠B = (3x - 5)°, ∠C = (4y)° and ∠D = (7y + 5)°.

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 y value in (1),

4x – 4(15) – 160 = 0

4x 220 = 0

x = 55

The solution is x = 55, y = 15.

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