In triangle PQR, points A, B and C are taken on PQ, PR and QR respectively such that QC = AC and CR = CB. If ∠QPR = 40°, then ∠ACB is equal to:

According to question,

In ΔPQR
x + y + 40° = 180°
x + y = 140° . . . . . . . (i)
In ΔAQC
x + x + ∠C = 180°
∠C = 180° - 2x . . . . . . . . (ii)
In ΔBCR
y + y + ∠C = 180°
∠C = 180° - 2y . . . . . . . . (iii)
But ∠ACB = 180° - 180° + 2x - 180° +2y
∠ACB = 2x + 2y - 180°
∠ACB = 2(x + y) - 180° . . . . . . . . . (iv)
Put the value of equation (i) in equation (iv)
∠ACB = 2 × 140° - 180°
∠ACB = 280° - 180°
∠ACB = 100°