Kąt α jest ostry oraz 3sinα-√(3)cosα=0. Wtedy

$\displaystyle \tg\alpha= \frac{1}{3}$\displaystyle \tg\alpha= \frac{1}{3}

$\tg\alpha=3$\tg\alpha=3

$\tg\alpha=\sqrt{3}$\tg\alpha=\sqrt{3}

$\displaystyle \tg\alpha= \frac{\sqrt{3}}{3}$\displaystyle \tg\alpha= \frac{\sqrt{3}}{3}