Q. 204.0( 4 Votes )

Find the angles marked with a question mark shown in Fig. 17.27


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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