Let A B C D E F G H I J K L be a regular dodecagon , then the value of A F A B + A B A F
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Sir nice approach . I am really inspired from you
W e w i l l s o l v e i n c o m p l e x p l a n e , θ = 1 2 2 π = 6 π , l e t A i s a t e i ( 0 ) B = e i 6 π , F = e i 6 5 π ⇒ A B = ∣ B − A ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ e i 6 π − 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 − 3 & A F = ∣ F − A ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ e i 6 5 π − 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 + 3 ∴ A F A B + A B A F = A B . A F A B 2 + A F 2 = 4 − 3 2 − 3 + 2 + 3 = 4
Angle half the side makes at the center for AB 15 degrees, and for AF=5 * 15=75.
So AB/AF=(2RSin15) / (2RSin75) +(2RSin15) / (2RSin75)= Tan15 + 1/Tan15=4. Why is it Level 5?
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Since ratio is independent of side length so let vertex as complex 12th roots of unity.This is simple application of complex numbers.
See this Problem solution Click here