According to the question,
ABCD is parallelogram,
DP ⊥ AB
DQ ⊥ BC.
∠PDQ = 60o
In quad. DPBQ,
Using angle sum property of a quadrilateral,
We have,
∠PDQ + ∠Q + ∠P + ∠B = 360o
60o + 90o + 90o + ∠B = 360o
240o + ∠B = 360o
∠B = 360o – 240o
∠B = 120o
Since, opposite angles in parallelogram are equal,
We have,
∠B = ∠D = 120o
Since, opposite sides are parallel in parallelogram,
We have,
AB||CD
Also, since sum of adjacent interior angles is 180o,
We have,
∠B + ∠C = 180o
120o + ∠C = 180o
∠C = 180o – 120o
∠C = 60o
Since, opposite angles in parallelogram are equal,
We have,
∠C = ∠A = 60o