Odpowiedź:
Szczegółowe wyjaśnienie:
Zad1.
Zaczynamy od obliczenia boku b uzywajac twierdzenia Pitagorasa:
[tex]a^2+b^2 = c^2\\12^2 + b^2 = 13^2\\b^2 = 169 - 144 = 25\\b = 5[/tex]
Nastepnie funkcje trygonometrycznie (kata poniedzy krawedziami a oraz c):
[tex]\sin\alpha = \frac{5}{13}\\\\\cos\alpha = \frac{12}{13}\\\\\tan\alpha = \frac{5}{12}\\\\\cot\alpha = \frac{12}{5}\\\\\\[/tex]
Zad.2
[tex]\sin^2 L +\cos^2 L = 1\\\sin^2 L = 1 - \cos^2 L = 1- (3/4)^2 = 1 - 9/16 = 7/16\\ | \sin L |= \sqrt{7/16} = \sqrt{7}/4\\\\ \sin L = \sqrt{7}/4 \quad albo \quad \sin L = - \sqrt{7}/4. \\\\\tan L = \frac{\sin L}{\cos L} = \frac{\sqrt{7}/4}{3/4} = \frac{\sqrt{7}}{3}\\albo\\\tan L = \frac{\sin L}{\cos L} = \frac{-\sqrt{7}/4}{3/4} = -\frac{\sqrt{7}}{3}.\\[/tex]
Zad. 3
Pole trojkata wynosi: [tex]P = 1/2 a b *\sin\alpha.[/tex]
[tex]P = 1/2 * 6 * 8 *\sin(45^{ \circ}\!) = 24 * \sqrt{2}/2 = 12\sqrt{2}.[/tex]
