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Bonjour,

a) 1° = 60' = 60 x 60" = 3600"

donc 1" = 1/3600 ° soit environ 2,77.10⁻⁴ °

b) tan(α) = ST/SR

c) tan(α) = d(T-S)/1 pc

⇒ 1 pc = d(T-S)/tan(α)

soit 1 pc ≈ (1,5.10¹¹)/tan(2,77.10⁻⁴) ≈ 3,1.10¹⁶ m

d) 1 al = 365 x 24 x 3600 x 3.10⁸ m ≈ 9,46.10¹⁵ m

⇒ 1 pc ≈ 3,1.10¹⁶/9,46.10¹⁵ ≈ 3,3 al  

(sans tenir compte des chiffres significatifs 1 pc ≈ 3,26 al)