Question

in rhombus abcd, what is the m∠abd if m∠bdc = 30°?

Explanation:

Step1: Recall rhombus properties

In a rhombus, the diagonals are perpendicular bisectors of each other and bisect the angles of the rhombus.

Step2: Identify related angles

Since $ABCD$ is a rhombus, $AD\parallel BC$. $\angle ADB=\angle BDC = 30^{\circ}$ (diagonal $BD$ bisects $\angle ADC$).

Step3: Use triangle - angle sum

In $\triangle ABD$, $\angle BAD = 180^{\circ}- 2\times30^{\circ}=120^{\circ}$ (because $\angle ADB=\angle ABD$ as $AD = AB$ in a rhombus). In $\triangle ABD$, using the angle - sum property of a triangle ($\angle ADB+\angle ABD+\angle BAD = 180^{\circ}$), and since $\angle ADB = 30^{\circ}$ and $\angle BAD=120^{\circ}$, we can find $\angle ABD$. Let $\angle ABD=x$. Then $30^{\circ}+x + 120^{\circ}=180^{\circ}$, so $x = 30^{\circ}$. And $\angle ABD = 60^{\circ}$ (because $\angle ABC$ is bisected by $BD$ and in rhombus adjacent angles are supplementary, and also using the isosceles - triangle property of $\triangle ABD$ where $AD = AB$).

Answer:

$60^{\circ}$