Question

ABCD is a rhombus, $\angle ADC={53}^{\circ }$. Diagonals AC and BD intersect at O. Find $\angle DAO$. [3 MARKS]

Answer 1

Concept: 1 Mark
Application: 2 Marks

In a rhombus, diagonals bisect the interior angles.

$\angle ADO=\frac{1}{2}\left(\angle ADC\right)$
$\angle ADO=\frac{1}{2}\times {53}^{\circ }$ [Substitute]
$\angle ADO={26.5}^{\circ }$

$\angle AOD={90}^{\circ }$ [Diagonals intersect at right angles]

In $△AOD$,
$\angle AOD+\angle ADO+\angle DAO={180}^{\circ }$ [Angle sum property]
$\Rightarrow {90}^{\circ }+{26.5}^{\circ }+\angle DAO={180}^{\circ }$
$\Rightarrow \angle DAO={180}^{\circ }-{90}^{\circ }-{26.5}^{\circ }={63.5}^{\circ }$