Answer:
[tex]108^o[/tex]
Step-by-step explanation:
Consider this regular pentagon [tex]ABCDE[/tex].
Let us join vertices [tex]AC[/tex] and [tex]EC[/tex] as shown to form three triangles as shown. I have used letters [tex]a,b,c,d,e,f,g,h,i[/tex] to represent internal angles of triangles for sake of simplicity.
Since the sum of interior angles of a triangle is [tex]180^o[/tex],
In △[tex]ABC,b+c+d=180^o[/tex]
In △[tex]ACE,a+e+180^o[/tex]
in △ [tex]ECD,h+f+g=180^o[/tex]
um of interior angles of the pentagon is
[tex]a+b+c+d+e+f+g+h+i[/tex]
[tex]=(b+c+d)+(a+e+i)+(h+f+g)[/tex]
[tex]=180^o + 180^o +1 80^o[/tex] [using the above three results]
[tex]=540^o[/tex]
[tex]i.e.[/tex] ∠A + ∠B + ∠C + ∠D + ∠E = [tex]540^o[/tex]
Since it is a regular octagon, ∠A = ∠B = ∠C = ∠D = ∠E
⇒ ∠A + ∠A + ∠A + ∠A + ∠A = [tex]540^o[/tex]
⇒ [tex]\iimplies 5*\angle{A} = 540^o[/tex]\
[tex]\implies \angle{A=540/5=108^o = \angle{B=\angle{C=\angle{D=\angle{E[/tex]
Hence internal angle of a regular pentagon is [tex]108^o[/tex]
