Question

and exterior angle of a parallelogram is 110 degree find the angles of the parallelogram

Answer 1

ABCD is a parallelogram.

∠CBX is exterior angle of parallelogram ABCD.

∠CBX +∠ABC  = 180°

110° +∠ABC  = 180°

∠ABC = 180° - 110°

∠ABC = 70°

∠B =∠D

[  Opposite angles of parallelogram are equal ]

so, ∠D = 70°

∠DCB +∠ABC  = 180°

[  Supplementary angles ]

∠DCB + 70° = 180°

∠DCB= 180° - 70°

∠DCB = 110°

Now,

∠C = ∠A

[  Opposite angles of parallelogram are equal ]

= 110°

Answer 2

Answer:

ABCD is a parallelogram.

∠CBX is exterior angle of parallelogram ABCD.

∠CBX +∠ABC  = 180°

110° +∠ABC  = 180°

∠ABC = 180° - 110°

∠ABC = 70°

∠B =∠D

[  Opposite angles of parallelogram are equal ]

so, ∠D = 70°

∠DCB +∠ABC  = 180°

[  Supplementary angles ]

∠DCB + 70° = 180°

∠DCB= 180° - 70°

∠DCB = 110