Question

In the given square ABCD, a line segment DE cuts the side BC at E and the diagonal AC at O such that $\angle COD={120}^{\circ }.$ Find the measure of $\angle OEC.$

• A. ${60}^{\circ }$
• B. ${65}^{\circ }$
• C. ${70}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is D ${75}^{\circ }$

$\text{Given:}\angle COD={120}^{\circ }$

$\text{We know that,}\angle EOC+\angle COD={180}^{\circ }\Rightarrow \angle EOC+{120}^{\circ }={180}^{\circ }\Rightarrow \angle EOC={60}^{\circ }$

$\text{We know that, the diagonals of a}$$\text{square bisect its angles}.$

$\text{Here, AC is the diagonal.}\text{Hence,}\angle OCE=\angle OED={45}^{\circ }$

$\text{In}△OCE$,
$\angle OCE+\angle EOC+\angle OEC={180}^{\circ }$
$\Rightarrow {45}^{\circ }+{60}^{\circ }+\angle OEC={180}^{\circ }$
$\Rightarrow \angle OEC={180}^{\circ }-{105}^{\circ }$
$\Rightarrow \angle OEC={75}^{\circ }$