Question
A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find ∠BCF?
A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find ∠BCF?
Answer:
:
Given, ABCDE is a regular pentagon.
Then, measure of each interior angle of the regular pentagon
=SumofinterioranglesNumberofsides=(x−2)×180∘5=(5−2)×180∘5=540∘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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:
Given, ABCDE is a regular pentagon.
Then, measure of each interior angle of the regular pentagon
=SumofinterioranglesNumberofsides=(x−2)×180∘5=(5−2)×180∘5=540∘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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