Question

given quad abcd is a rhombus and m∠abd = 32°. find the measure of each numbered angle.

Explanation:

Step1: Properties of a rhombus

In a rhombus, the diagonals bisect the vertex angles. So, \(\angle ABD=\angle1 = 32^{\circ}\) (diagonal \(BD\) bisects \(\angle ABC\)).

Step2: Right - angled property of rhombus diagonals

The diagonals of a rhombus are perpendicular to each other. So, \(\angle2=90^{\circ}\).

Step3: Triangle angle - sum property

In \(\triangle BOC\) (where \(O\) is the intersection of the diagonals), using the triangle angle - sum property (\(\angle1+\angle2+\angle3 = 180^{\circ}\)). Substitute \(\angle1 = 32^{\circ}\) and \(\angle2=90^{\circ}\). Then \(32^{\circ}+90^{\circ}+\angle3=180^{\circ}\), so \(\angle3=180^{\circ}-(32^{\circ} + 90^{\circ})=58^{\circ}\).

Step4: Diagonal bisecting property

The diagonals of a rhombus bisect the vertex angles. Diagonal \(BD\) bisects \(\angle ADC\), so \(\angle4=\angle ABD = 32^{\circ}\) (alternate interior angles for parallel sides \(AB\parallel CD\) and transversal \(BD\)).

Answer:

\(\angle1 = 32^{\circ}\), \(\angle2=90^{\circ}\), \(\angle3 = 58^{\circ}\), \(\angle4=32^{\circ}\)