Question

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

Answer 1

Given: A = (4x + 20)°, B = (3x – 5)°, C = (4y)°, D = (7y + 5)°.

In a cyclic quadrilateral, the sum of each pair of opposite angles is 180°.

Step-wise calculation:

1. A + C = 180°

(4x + 20) + 4y = 180

4x + 4y + 20 = 180

4x + 4y = 160

x + y = 40

2. B + D = 180°

(3x – 5) + (7y + 5) = 180 

3x + 7y = 180

3. Solve the system:

From x + y = 40 ⇒ x = 40 – y.

Substitute in 3x + 7y = 180:

3(40 – y) + 7y = 180

120 – 3y + 7y = 180

120 + 4y = 180

4y = 60

⇒ y = 15

Then x = 40 – 15

= 25

4. Compute angles:

∠A = 4x + 20 

= 4(25) + 20

= 100 + 20

= 120°

∠B = 3x – 5

= 3(25) – 5

= 75 – 5

= 70°

∠C = 4y

= 4(15) 

= 60°

∠D = 7y + 5

= 7(15) + 5

= 105 + 5

= 110°

The angles are ∠A = 120°, ∠B = 70°, ∠C = 60° and ∠D = 110°.