Question

A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find ∠BCF.

Answer 1

Given, ABCDE is a regular pentagon.

Then, measure of each interior angle of the regular pentagon

= `"Sum of interior angles"/"Number of sides"`

= `((x - 2) xx 180^circ)/5`

= `((5 - 2) xx 180^circ)/5`

= `540^circ/5`

= 108°

∴ ∠CBA = 108°

Join CF,

Now, ∠FBC = 360° – (90° + 108°)

= 360° – 198°

= 162°

In ΔFBC, by the angle sum property, we have

∠FBC + ∠BCF + ∠BFC = 180°

⇒ ∠BCF + ∠BFC = 180° – 162°

⇒ ∠BCF + ∠BFC = 18°

Since, ΔFBC is an isosceles triangle and BF = BC.

∴ ∠BCF = ∠BFC = 9°