Find the angles marked with a question mark shown in Fig.
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In ΔBEC
∠BEC + ∠ECB +∠CBC = 180° [Sum of angles of a triangle is 180°]
90° + 40° + ∠CBC = 180°
∠CBC = 180°-130°
∠CBC =50°
∠B = ∠D = 50° [Opposite angles of a parallelogram are equal]
∠A + ∠B = 180° [Sum of adjacent angles of a triangle is 180°]
∠A + 50° = 180°
∠A = 180°- 50°
∠A = 130°
In ΔDFC
∠DFC + ∠FCD +∠CDF = 180° [Sum of angles of a triangle is 180°]
90° + ∠FCD + 50° = 180°
∠FCD = 180°- 140°
∠FCD =40°
∠A = ∠C = 130° [Opposite angles of a parallelogram are equal]
∠C = ∠FCE +∠BCE + ∠FCD
∠DCF + 40° + 40° = 130°
∠DCF = 130°- 80°
∠DCF = 50°