Question

ABCD is a rectangle with angle BAC=32°. determine angle DBC

Answer 1

If ABCD IS A PARALLELOGRAM NOT A RECTANGLE, THEN
Angle BAC + DBA = 180° ( consecutive interior angle)
= 32° + DBA = 180°
= DBA = 180° - 32° = 148°

NOW, we know that the diagonal of the parallelogram divides the angles into two equal parts.
so, ANGLE DBA = DBC + ABC
= 148° = 2DBC. ( DBC AND ABC ARE EQUAL)
= DBC = 74° ANS

Answer 2

AnswEr:

Suppose the diagonals AC and AB intersect at O.

• $\sf\underline{\:\:\:In\:\:\:\triangle\:OAB,\:\:\:we\:\:\:have\:\:\::-}$

$\\ \\ \sf \: OA = OB \qquad( \because \: diagonals \: of \: a \: rectangle \: are \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: equal \: and \: bisect \: each \: other) \\ \\ \\ \implies \angle \sf \: OAB = \angle \: OBA \\ \\ \implies \sf \angle \: BAC = \angle \: DBA \\ \\ \implies \sf \angle \: DBA = 32 \degree \qquad \: ( \because \angle \: BAC = 32 \degree)$

___________________

Now,

$\qquad \sf \angle \: ABC = 90 \degree \\ \\ \implies \sf \angle \: DBA + \angle \: DBC = 90 \degree \\ \\ \implies \sf \: 32 \degree + \angle \: DBC = 90 \degree \\ \\ \implies \sf \angle \: DBC = 90 \degree - 32 \degree = 58 \degree$