Question

an exterior angles of parallelogram is 110 degree find angles of the
parallelogram

Answer 1

This is the answer for your answer .I hope it is useful for you.

Answer 2

Step-by-step explanation:

SOLUTION :

ABCD is a parallelogram.

\angle{CBX}∠CBX is exterior angle of parallelogram ABCD.

\angle{CBX}∠CBX + \angle{ABC}∠ABC = 180°

110° + \angle{ABC}∠ABC = 180°

\angle{ABC}∠ABC = 180° - 110°

\angle{ABC}∠ABC = 70°

\angle{B}∠B = \angle{D}∠D

[ \therefore∴ Opposite angles of parallelogram are equal ]

so, \angle{D}∠D = 70°

\angle{DCB}∠DCB + \angle{ABC}∠ABC = 180°

[ \therefore∴ Supplementary angles ]

\angle{DCB}∠DCB + 70° = 180°

\angle{DCB}∠DCB = 180° - 70°

\angle{DCB}∠DCB = 110°

Now,

\angle{C}∠C = \angle{A}∠A

[ \therefore∴ Opposite angles of parallelogram are equal ]

\angle{A}∠A = 110°

hope it will help u ❤️