Question

find:
FAB+ABC+BCD+CDE+DEF+EFA ....

plz don't answer if u don't know the answer.... and don't u dare spam my question.... ​

Answer 1

Answer:

In ΔABC, we have

∠BAC + ∠ABC + ∠BCA = 180º ..... (i)

In ΔADC, we have

∠CAD + ∠ADC + ∠ACD = 180º ..... (ii)

In ΔADE, we have

∠DAE + ∠ADE + ∠DEA = 180º ..... (iii)

In ΔAEF, we have

∠EAF + ∠AEF + ∠AFE = 180º ..... (iv)

Adding (i), (ii), (iii), (iv) and regrouping, we get

(∠BAC + ∠CAD + ∠DAE + ∠EAF) + ∠ABC + (∠BCA + ∠ACD) + (∠ADC + ∠ADE)

+ (∠DEA + ∠AEF) + ∠AFE = 720º

⇒ ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720º

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Answer 2

Answer:

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