In Fig. 16.21, th

Sum of angles of a quadrilateral is 360°

In the quadrilateral ABCD

∠D = 50°, ∠C = 100°,

∠A + ∠B + ∠C + ∠D = 360°

∠A + ∠B + 100° + 50° =

∠A + ∠B =

∠A + ∠B = 210° ……….(i)

Now in Δ APB

[Sum of angles of a triangle is 180°]

…….(ii)

On substituting value of ∠A + ∠B = 210 from equation (i) in equation (ii)