1.

Find the angles marked with a question mark shown in Fig.

Answer»

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°


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