Question

In the above figure, ABCD is a square and BCE is an equilateral triangle. Find the measure of angle DEC.

Answer 1

Given ABCD is a square and BCE is an equilateral triangle

$\angle BCD={90}^{\circ }$ [Interior angle of a square]

$\angle BCE={60}^{\circ }$ [Interior angle of an equilateral triangle]

$\therefore \angle DCE={90}^{\circ }+{60}^{\circ }={150}^{\circ }$

BC = CD (Sides of a square) ------ (1)

BC = CE (Sides of an equilateral triangle) ------ (2).

From (1) and (2),
DC = EC

In $\Delta DCE$, DC = CE
$\Rightarrow \angle CDE=\angle CED$ --------- (3) [Angles opposite equal sides]

Also, In $\Delta DCE$
$\angle DCE+\angle CDE+\angle CED={180}^{\circ }$ --------- (4) [Angle sum property]

From (3) and (4),
$2\angle DEC+\angle CED={180}^{\circ }$

$\Rightarrow 2\angle DEC={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

$\angle DEC=\frac{1}{2}\left({30}^{\circ }\right)={15}^{\circ }$