Question

Question 182
A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find $\angle BCF$?

Answer 1

Given, ABCDE is a regular pentagon.
Then, measure of each interior angle of the regular pentagon
$=\frac{Sumofinteriorangles}{Numberofsides}=\frac{(x-2)\times {180}^{\circ }}{5}=\frac{(5-2)\times {180}^{\circ }}{5}=\frac{{540}^{\circ }}{5}={108}^{\circ }\therefore \angle CBA={108}^{\circ }$
Join CF.
Now, $\angle FBC={360}^{\circ }-({90}^{\circ }+{108}^{\circ })={360}^{\circ }-{198}^{\circ }={162}^{\circ }$
In $\Delta FBC$, by the angle sum property, we have
$\angle FBC+\angle BCF+\angle BFC={180}^{\circ }\Rightarrow \angle BCF+\angle BFC={180}^{\circ }-{162}^{\circ }\Rightarrow \angle BCF+\angle BFC={18}^{\circ }$
Since, $\Delta FBC$ is an isosceles triangle and BF = BC.
$\therefore \angle BCF=\angle BFC=9^{\circ }$