Question

An exterior angle of a parallelogram is 110°. Find the angles of the parallelogram

Answer 1

$\mathbb{ SOLUTION }$:

ABCD is a parallelogram.

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

$\angle{CBX}$ + $\angle{ABC}$ = 180°
110° + $\angle{ABC}$ = 180°
$\angle{ABC}$ = 180° - 110°
$\angle{ABC}$ = 70°

$\angle{B}$ = $\angle{D}$
[ $\therefore$ Opposite angles of parallelogram are equal ]
so, $\angle{D}$ = 70°

$\angle{DCB}$ + $\angle{ABC}$ = 180°
[ $\therefore$ Supplementary angles ]

$\angle{DCB}$ + 70° = 180°
$\angle{DCB}$ = 180° - 70°
$\angle{DCB}$ = 110°

Now,
$\angle{C}$ = $\angle{A}$
[ $\therefore$ Opposite angles of parallelogram are equal ]
$\angle{A}$ = 110°

Answer 2

Here is your answer

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