[Risolto] aiuto mate

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$AC= 8tan\big[cos^{-1}\big(\frac{4}{5}\big)\big]=8×tan(36,869898°)=6$;

$tan(γ)= tan\big[90°-cos^{-1}\big(\frac{4}{5}\big)\big]= \frac{4}{3}$.

se cos β = 4/5 (0,800) , allora sen β = 0,600

BC = AB/cos β = 8/0,8 = 10 cm

AC = BC*sen β = 10*0,6 = 6 cm

tan γ = AB/AC = 0,8/0,6 = 4/3