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

1.	Find the angles marked with a question mark shown in Figure.
Answer» In ΔBEC ∠BEC + ∠ECB +∠CBE = 180° [Sum of angles of a triangle is 180°] 90° + 40° + ∠CBE = 180° ∠CBE = 180°-130° ∠CBE = 50° ∠CBE = ∠ADC = 50° (Opposite angles of a parallelogram are equal) ∠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° So, ∠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 ∠FCD + 40° + 40° = 130° ∠FCD = 130° – 80° ∠FCD = 50° ∴ ∠EBC = 50^o, ∠ADC = 50^o and ∠FCD = 50^o

Answer»

In ΔBEC

∠BEC + ∠ECB +∠CBE = 180° [Sum of angles of a triangle is 180°]

90° + 40° + ∠CBE = 180°

∠CBE = 180°-130°

∠CBE = 50°

∠CBE = ∠ADC = 50° (Opposite angles of a parallelogram are equal)

∠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°

So, ∠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

∠FCD + 40° + 40° = 130°

∠FCD = 130° – 80°

∠FCD = 50°

∴ ∠EBC = 50^o, ∠ADC = 50^o and ∠FCD = 50^o

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