Question

In the given diagram, equilateral triangle $EAB$ surmounts square $ABCD$. Find the the measure of $\angle AEC$.

• A. ${30}^{\circ }$
• B. ${15}^{\circ }$
• C. ${45}^{\circ }$
• D. ${25}^{\circ }$

Answer 1

The correct option is C ${45}^{\circ }$

In $△BCE$
$\angle BCE+\angle BEC+\angle CBE={180}^{\circ }$
$[BE=BC⟹\angle BCE=\angle BEC]$
And,
Since $△EAB$ is equilateral and ABCD is square,
$\angle CBE={90}^{\circ }+{60}^{\circ }={150}^{\circ }$
$⟹2\angle BEC+{150}^{\circ }={180}^{\circ }⟹\angle BEC={15}^{\circ }$
Since $△EAB$ is equilateral,
$\angle AEB={60}^{\circ }⟹\angle AEC+\angle BEC={60}^{\circ }⟹\angle AEC={60}^{\circ }-{15}^{\circ }={45}^{\circ }$