Question

In the above figure, ABCD is a square and triangle ADE is equilateral. Find $\angle EFD$.

• A. ${65}^{\circ }$
• B. ${70}^{\circ }$
• C. ${80}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

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

Given that:

$ABCD$ is a square and $ADE$ is an equilateral triangle.

To find:

$\angle EFD=?$

Solution:

$\angle EAB=\angle EAF+\angle BAF$

or, $\angle EAB={60}^{\circ }+{90}^{\circ }={150}^{\circ }$

Since, $AE=AB$

$\therefore \angle AEB=\angle ABE$

In $△AEB,$

$\angle AEB+\angle ABE+\angle EAB={180}^{\circ }$

or, $2\angle AEB={180}^{\circ }-{150}^{\circ }$

or, $\angle AEB={15}^{\circ }$

Now,

$\angle EFD=\angle EAF+\angle AEF$ $(\angle AEF=\angle AEB)$

(Sum of remote interior angles is equal to exterior angle)

or, $\angle EFD={60}^{\circ }+{15}^{\circ }$

or, $\angle EFD={75}^{\circ }$

Hence, D is the correct option.