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Home » z = 30^\circ, \quad z = 60^\circ - Simpleprint

z = 30^\circ, \quad z = 60^\circ - Simpleprint

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 u = rac{3 \pm \sqrt{9 - 4}}{2} = rac{3 \pm \sqrt{5}}{2}.

📌 So \( u = rac{3 + \sqrt{5}}{2} \) or \( u = rac{3 - \sqrt{5}}{2} \).

📌 Now recall \( u = rac{x+y}{x-y} \). Let \( r = rac{x}{y} \), assuming \(y

📌 e 0\). Then:

📌 rac{x+y}{x-y} = rac{r+1}{r-1}.

📌 Set this equal to each value of \(u\):

📌 Case 1:** \( rac{r+1}{r-1} = rac{3 + \sqrt{5}}{2} \)

📌 2(r+1) = (3 + \sqrt{5})(r - 1)

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L \equiv 0 \pmod{7} \
Use the Chinese Remainder Theorem. Let $ L = 7k $. Substitute into the...
7k \equiv 1 \pmod{13}
We solve for $ k $: find the modular inverse of 7 modulo 13.
$ 7 \cdot 1 = 7
$ 7 \cdot 2 = 14 \equiv 1 \pmod{13} $
So $ 7^{-1} \equiv 2 \pmod{13} $
k \equiv 2 \pmod{13} \Rightarrow k = 13m + 2
Then $ L = 7k = 7(13m + 2) = 91m + 14 $
For $ m = 0 $, $ L = 14 $ (two-digit)

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