Question

In the given figure : ABCD is a rhombus with angle $A={67}^{\circ }$.
If DEC is an equilateral triangle, calculate $\angle CBE$

Answer 1

$GivenisarhombusABCDwith\angle A={67}^o\&△DECisanequilateraltriangle.SoDE=EC=CD\&\angle DEC=\angle ECD=\angle CDE={60}^{o}AlsoEC=CD=CB\left[Sinceallsidesoftherhombusareequal\right]Now\angle A=\angle BCD\left[Oppositeanglesareequalinarhombus\right]So\angle ECB=\angle ECD+\angle BCD\Rightarrow \angle ECB={67}^o+{60}^o\Rightarrow \angle ECB={127}^{o}SinceEC=CBso△ECBisisoscelestriangle.In△ECB,\angle ECB+\angle BEC+\angle CBE={180}^o$$\Rightarrow 2\angle CBE+{127}^0={180}^0[Baseanglesinisosceles△areequal\Rightarrow \angle CEB=\angle CBE]\Rightarrow 2\angle CBE={180}^0-{127}^0\Rightarrow 2\angle CBE={53}^0\Rightarrow \angle CBE=\frac{{53}^0}{2}$

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