In the diagram shown, ^@ ABCD ^@ is a square and point ^@ F ^@ l | Math Question and Answer

Answer:

$75^\circ$

Step by Step Explanation:

Given, $ABCD$ is a square, $\triangle DEC$ is an equilateral triangle and $EB =EF.$
$\implies \angle DCB = 90^\circ \text{ and } \angle DCE = 60^\circ$
$\implies \angle ECF = 30^\circ$
Since $DC = CE \space\space\space\space$ [Sides of an equilateral triangle]
and $DC = CB \space\space\space\space$ [Sides of a square]
$\implies CE = CB$
$\implies \triangle ECB$ is an isosceles triangle.
$\implies \angle EBC = \angle BEC\space\space\space\space\space\space\space\space\space\space\space\space [\because \text{Angles opposite to equal sides of a triangle are equal}] \space\space\space\space$
Now, $\angle ECB + \angle EBC + \angle BEC = 180^\circ \space\space\space\space\space\space\space\space\space\space\space\space [\text{ Angle Sum Property of a triangle}]$
$\implies \angle EBC + \angle EBC + 30^\circ = 180^\circ$
$\implies \angle EBC = \dfrac{ (180 - 30) } { 2 } = 75^\circ$
Given, $EB = EF$
$\therefore \angle BFE = \angle EBC = 75^\circ$
Hence, the value of $\angle EBC$ is $\angle 75^\circ.$

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