Answer:
m∠A 60°, m∠B= 120°, m∠C =60°, m∠D=120°
Step-by-step explanation:
[tex] In\: ||^{gm} \: ABCD, \: BK\perp AD... (given) \\
\therefore \angle BKA = 90\degree \\
In\:\triangle BKA, \\\\
\cos \angle A = \frac{AK}{AB}\\\\
\cos \angle A = \frac{3}{6}\\\\
\cos \angle A = \frac{1}{2}\\\\
\cos \angle A = \cos 60\degree.. (\because \cos 60\degree = \frac{1}{2}) \\\\
\huge \red {\boxed {\therefore \angle A = 60\degree}} \\[/tex]
Since, adjacent angles of a parallelogram are Supplementary.
[tex] \therefore \angle A + \angle B = 180\degree \\
\therefore 60\degree + \angle B = 180\degree \\
\therefore \angle B = 180\degree - 60\degree\\
\huge \purple {\boxed {\therefore \angle B = 120\degree}} \\[/tex]
Since, opposite angles of a parallelogram are congruent.
[tex] \therefore \angle C = \angle A
\\
\huge \red{\boxed {\therefore \angle C = 60\degree}} \\\\
\therefore \angle D = \angle B
\\
\huge \purple {\boxed {\therefore \angle D = 120\degree}} \\[/tex]
