Question

Let $ABCD$ is a rhombus with angle $A={67}^0$. If $DEC$ is an equilateral
triangle, then $\angle CBE$ is equal to ${26.5}^0={26}^0{30}^{\prime }$

State true or false:

• A. True
• B. False

Answer 1

The correct option is A True

Given is a rhombus ABCD with $\angle A={67}^o$ & $△$ 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,

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

$\Rightarrow 2\angle CBE+{127}^o={180}^o$[ Base angles in isosceles triangle are equal $\Rightarrow \angle CEB=\angle CBE$ ]

$\Rightarrow 2\angle CBE={180}^o-{127}^o$

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

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

$\Rightarrow \angle CBE={26.5}^o$