Question

The diagonals of a rectangle $ABCD$ meet at $O$. If $\angle BOC={44}^{\circ }$, find $\angle OAD$. [2 MARKS]

Answer 1

Concept : 1 Mark
Calculation : 1 Mark

In rectangle $ABCD$,

$\angle AOD={44}^{\circ }$ [vertically opposite]

$\angle ODA=\angle OAD=x^{\circ }$ [Since $\Delta OAD$ is an isosceles triangle)

$\therefore$ By the angle sum property of a triangle, we have
$\Rightarrow \angle OAD$ + $\angle ODA$ + $\angle AOD$ = ${180}^{\circ }$
$\Rightarrow x^{\circ }+x^{\circ }+{44}^{\circ }={180}^{\circ }$
$\Rightarrow 2x^{\circ }+{44}^{\circ }={180}^{\circ }$
$\Rightarrow x^{\circ }=\frac{{180}^{\circ }-{44}^{\circ }}{2}=\frac{{136}^{\circ }}{2}={68}^{\circ }$
$\therefore \angle OAD={68}^{\circ }$