dalam ΔABC jika AB=6 AC=8 dan tan ∠ABC=2[tex]\sqrt3 [/tex] maka tab∠BAC= ... mohon dibantu ya kaa makasih :)

Tan mi = √((2√3)^2 + 1^2) = √(12 + 1) = √13
Sin B = de/mi = (2√3)/√13

AB/(sin C) = AC/(sin B)
6/(sin C) = 8/(2√3/√13)
6/(sin C) = (8√13)/(2√3)
6/(sin C) = (4√13)/√3
3/(sin C) = (2√13)/√3
2√13 sin C = 3√3
Sin C = (3√3)/(2√13) = de/mi
sa = √((2√13)^2 - (3√3)^2) = √(52 - 27) = √25 = 5
Tan C = de/sa = (3√3)/5

A + B + C = 180° => A = (180° - (B + C))
Tan <BAC
= Tan A
= Tan (180° - (B + C))
= - Tan (B + C)
= - (Tan B + Tan C)/(1 - Tan B Tan C)
= - (2√3 + (3√3)/5) / (1 - 2√3 .(3√3)/5) . 5/5
= - (10√3 + 3√3) / (5 - 18)
= - (13√3) / -13
= √3