Question

In the figure above (not to scale) ABCD is a rectangle with diagonals intersecting at O, If OA = AB and $\angle AEC={120}^{\circ }$ then $\angle OAF$ = ______

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${50}^{\circ }$
• D. Cannot be determined

Answer 1

The correct option is A ${30}^{\circ }$

Since the given figure is a rectangle so, $OB=OA$ and it is given that $OA=AB$

Therefore, $OA=OB=AB$

Hence, triangle $OAB$ is an equilateral triangle.

$\angle OAB=$ ${60}^{\circ }$

$\angle AEC={120}^{\circ };\angle AEB={60}^{\circ }$ {Sum of linear Pair angles $={180}^o$}

$\angle EBA={90}^{\circ }$ {Property of Rectangle}

In $△EAB$,

$\angle EAB+\angle EBA+\angle AEB=\angle EAB+{90}^o+{60}^o={180}^{\circ }$ {Sum of angles of Triangle}

$\angle EAB={30}^{\circ }$

$\angle OAF=\angle OAB-\angle EAB$

$=(60-30{)}^o$

$={30}^{\circ }$

So correct answer will be Option A