Question

In the given figure : ABCD is a rhombus with angle $A={67}^{\circ }$.

If DEC is an equilateral triangle, calculate $\angle DBE$(in degrees).

Answer 1

ABCD is a rhombus with $\angle A={67}^o$ & $\Delta DEC$ is an equilateral triangle.

So DE= EC = CD & $\angle DEC=\angle ECD=\angle CDE={60}^o$

Also EC = CD = CB ...[ Since all sides of the rhombus are equal

Now, $\angle A=\angle BCD$ ...[Opposite angles are equal in a rhombus]

So $\angle ECB=\angle ECD+\angle BCD\Rightarrow \angle ECB={67}^o+{60}^o\Rightarrow \angle ECB={127}^o$

Since EC = CB, so $△ECB$ is isosceles triangle.

In $△ECB$,

$\angle ECB+\angle BEC+\angle CBE={180}^o\Rightarrow 2\angle CBE+127=180$

{Base angles in isosceles triangle are equal $\Rightarrow \angle CEB=\angle CBE$]

$\Rightarrow 2\angle CBE=180-127\Rightarrow 2\angle CBE=53$

$\Rightarrow \angle CBE=\frac{53}{2}\Rightarrow \angle CBE={26.5}^o$

In isosceles $△DAB,\angle ADB=\angle ABD$

[Base angles in isosceles triangle are equal

Again in isosceles $△DAB,\angle ADB+\angle ABD+\angle BAD={180}^o$

$\Rightarrow 2\angle ABD=180-67\Rightarrow 2\angle ABD={113}^o$

$\Rightarrow \angle ABD={56.5}^o$

In rhombus ABCD, adjacent angles are supplementary,

$\therefore \angle DAB+angleABC={180}^o\Rightarrow \angle ABC=180-67\Rightarrow \angle ABC={113}^o$

So $\angle DBE=\angle ABC-\angle ABD-\angle CBE$

$\Rightarrow \angle DBE={113}^o-{56.5}^o-{26.5}^o$

$\Rightarrow \angle DBE={113}^o-83$ $\Rightarrow \angle DBE={30}^o$