Question

Triunghiul isoscel ABC are AB=AC=12cm și m(<BAC)=45°. Calculează tg(<ABC).​

Answer 1

Explicație pas cu pas:

AD ⊥ BC; BC = 2×BD

$BC²=AB²+AC²-2×AB×AC×cos(BAC) = 2 \times {12}^{2} - 2 \times {12}^{2} \times \frac{ \sqrt{2} }{2} = {12}^{2} \times (2 - \sqrt{2})$

$aria(ABC)= \frac{AB×AC×sin(BAC)}{2} = \frac{AD×BC}{2} = > AD = \frac{AB×AC×sin(BAC)}{BC}$

$AD = \frac{ {12}^{2}× \frac{ \sqrt{2} }{2} }{BC} = \frac{ {12}^{2} \sqrt{2}}{2BC}$

$\tan(ABC) = \frac{AD}{BD} = \frac{AD}{ \frac{BC}{2} } = \frac{2AD}{BC} = \frac{ 2 \times {12}^{2} \sqrt{2}}{2BC^{2} } = \frac{ {12}^{2} \sqrt{2} }{ {12}^{2} (2 - \sqrt{2} )} = \frac{ \sqrt{2} }{2 - \sqrt{2} } = \frac{2( \sqrt{2} + 1) }{2} = \sqrt{2} + 1$

$\tan(ABC) = \sqrt{2} + 1$