Question

Given: ABCD ∥gram BK ⊥ AD , AB = 6, AK = 3 Find: m∠A, m∠B, m∠C, m∠D

Answer 1

m∠A 60°, m∠B= 120°, m∠C =60°, m∠D=120°

Explanation:

`In\: ||^(gm) \: ABCD, \: BK\perp AD... (given) \\</p><p>\therefore \angle BKA = 90\degree \\</p><p>In\:\triangle BKA, \\\\</p><p>\cos \angle A = (AK)/(AB)\\\\</p><p>\cos \angle A = (3)/(6)\\\\</p><p>\cos \angle A = (1)/(2)\\\\</p><p>\cos \angle A = \cos 60\degree.. (\because \cos 60\degree = (1)/(2)) \\\\</p><p>\huge \red {\boxed {\therefore \angle A = 60\degree}} \\`

Since, adjacent angles of a parallelogram are Supplementary.

`\therefore \angle A + \angle B = 180\degree \\</p><p>\therefore 60\degree + \angle B = 180\degree \\</p><p>\therefore \angle B = 180\degree - 60\degree\\</p><p>\huge \purple {\boxed {\therefore \angle B = 120\degree}} \\`

Since, opposite angles of a parallelogram are congruent.

`\therefore \angle C = \angle A</p><p> \\ </p><p>\huge \red{\boxed {\therefore \angle C = 60\degree}} \\\\</p><p></p><p>\therefore \angle D = \angle B</p><p> \\ </p><p>\huge \purple {\boxed {\therefore \angle D = 120\degree}} \\`