Question

In a cyclic quadrilateral ABCD, ∠A = (6x + 10)°, ∠B = (5x)°, ∠C = (x + y)° and ∠D = (3y – 10)°. What will be the four angles of the cyclic quadrilateral?

Answer 1

Given: ∠A = (6x + 10)°, ∠B = (5x)°, ∠C = (x + y)° and ∠D = (3y – 10)°

The sum of opposite angles in a cyclic quadrilateral equals 180°.

∴ ∠A + ∠C = 180° and ∠B + ∠D = 180°

Now, (6x + 10)° + (x + y)° = 180°

⇒ 7x + y = 180° – 10°

⇒ 7x + y = 170°  ......(i)

Also, (5x)° + (3y – 10)° = 180°

⇒ 5x + 3y = 180° + 10°

⇒ 5x + 3y = 190°  ......(ii)

Multiplying equation (i) by 3, we get

21x + 3y = 510°  .....(iii)

Now, subtracting equation (ii) from (iii), we get

21x + 3y = 510°
  5x + 3y = 190°
–     –      –         
        16x = 320°

⇒ x = 20°

Substituting the value of x in equation (ii), we get

5(20°) + 3y = 190°

100° + 3y = 190°

3y = 190° – 100° = 90°

y = 30°

So, x = 20° and y = 30°

Thus, ∠A = (6x + 10)° = (6 × 20 + 10)° = 130°

∠B = (5x)° = (5 × 20)° = 100°

∠C = (x + y)° = (20° + 30°) = 50°

∠D = (3y – 10)° = (3 × 30 – 10)° = 80°

Hence, the angles are 130°, 100°, 50° and 80°.