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