Question

In a cyclic quadrilateral ABCD, it is given that ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)° and ∠D = (4x – 5)°. Find the four angles.

Answer 1

Given: ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, ∠D = (4x – 5)° in cyclic quadrilateral ABCD.

Step-wise calculation:

1. In a cyclic quadrilateral opposite angles are supplementary, so ∠A + ∠C = 180° and ∠B + ∠D = 180°.

2. Write the equations:

(2x + 4) + (2y + 10) = 180

⇒ 2x + 2y + 14 = 180

⇒ 2x + 2y = 166

⇒ x + y = 83   ...(Equation 1)

(y + 3) + (4x – 5) = 180

⇒ 4x + y – 2 = 180

⇒ 4x + y = 182   ...(Equation 2)

3. Subtract equation 1 from equation 2:

(4x + y) – (x + y) = 182 – 83

⇒ 3x = 99 

⇒ x = 33

4. Substitute x = 33 into Eq. 1:

33 + y = 83

⇒ y = 50

5. Compute the angles:

∠A = 2x + 4

= 2(33) + 4 

= 66 + 4

= 70°

∠B = y + 3

= 50 + 3 

= 53°

∠C = 2y + 10

= 2(50) + 10 

= 100 + 10

= 110°

∠D = 4x – 5

= 4(33) – 5 

= 132 – 5 

= 127°

6. Check: ∠A + ∠C = 70° + 110° = 180°

And ∠B + ∠D = 53° + 127° = 180°   ...(Consistent)

∠A = 70°, ∠B = 53°, ∠C = 110°, ∠D = 127°.