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A regular pentagon $ABCDE$ and a square $ABFG$ are formed on opposite sides of $AB$. Find $\angle BCF$.
Given: A regular pentagon $ABCDE$ and a square $ABFG$ are formed on opposite sides of $AB$.
To do: To find $\angle BCF$.
Solution:
As given, $ABCDE$ is a regular pentagon and $ABFG$ is a square.
So the sides of pentagon and square are equal.
Each angle of square $=90^o$
Each angle of regular pentagon$=108^o$
From the figure $\angle FBC=360-( \angle ABF+\angle ABC)$
$\Rightarrow \angle FBC = 360-( 90+108)$
$\Rightarrow \angle FBC=360-198=162^o$
In $\vartriangle BCF$,
$\because BF=FC$
$\therefore BFC=BCF$
$\Rightarrow \angle BCF+\angle BCF+162^o=180^o$
$\Rightarrow 2\angle BCF=180^o-162^o$
$\Rightarrow \angle BCF=\frac{18^o}{2}$
$\Rightarrow \angle BCF=9^o$
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