Question

$\text{In given figure, ABCD is a rhombus.}\text{If}\angle OAB={40}^{\circ },\text{then find the value of x.}$

Answer 1

$\text{Given:}\angle OAB={40}^{\circ }$

$\text{We know that diagonals of a rhombus are}\text{perpendicular to each other.}\therefore \angle AOB={90}^{\circ }$

$\text{In}△AOB$,
$\text{By angle sum property,}\angle OAB+\angle AOB+\angle OBA={180}^{\circ }{40}^{\circ }+{90}^{\circ }+x={180}^{\circ }$
$x={180}^{\circ }-{130}^{\circ }$
$x={50}^{\circ }$