Odpowiedź:
Zad.1
Z rysunku wnioskujemy, że:
[tex]sin\alpha=\frac{3}{5}[/tex]
[tex]sin^{2} \alpha +cos^{2} \alpha =1[/tex]
[tex](\frac{3}{5})^{2}+cos^{2}\alpha=1[/tex]
[tex]\frac{9}{25}+cos^{2}\alpha =1[/tex]
[tex]cos^{2}\alpha=1-\frac{9}{25}=\frac{16}{25}[/tex]
[tex]cos\alpha =\sqrt{\frac{16}{25} }=\frac{4}{5}[/tex]
[tex]tg\alpha =\frac{sin\alpha }{cos\alpha }=\frac{\frac{3}{5} }{\frac{4}{5} }=\frac{3}{5}*\frac{5}{4}=\frac{3}{4}[/tex]
[tex]ctg\alpha = \frac{1}{tg\alpha } = \frac{1}{\frac{3}{4} }= 1*\frac{4}{3}= \frac{4}{3}[/tex]
Zad.2
Z rysunku wnioskujemy, że:
[tex]cos\alpha =\frac{6}{8}=\frac{3}{4}[/tex]
[tex]sin^{2}\alpha + cos^{2}\alpha = 1[/tex]
[tex]sin^{2}\alpha +(\frac{3}{4})^{2}=1[/tex]
[tex]sin^{2}\alpha +\frac{9}{16}=1[/tex]
[tex]sin^{2} \alpha = 1-\frac{9}{16}=\frac{7}{16}[/tex]
[tex]sin\alpha =\sqrt{\frac{7}{16} }=\frac{\sqrt{7} }{4}[/tex]
[tex]tg\alpha =\frac{sin\alpha }{cos\alpha } = \frac{\frac{\sqrt{7} }{4} }{\frac{3}{4} }=\frac{\sqrt{7} }{4}*\frac{4}{3}=\frac{\sqrt{7} }{3}[/tex]
[tex]ctg\alpha = \frac{1}{tg\alpha } = \frac{1}{\frac{\sqrt{7} }{3} }=1*\frac{3}{\sqrt{7} }*\frac{\sqrt{7} }{\sqrt{7} }=\frac{3\sqrt{7} }{7}[/tex]
[tex]\frac{sin\alpha -cos\alpha }{tg\alpha+ctg\alpha } = \frac{\frac{\sqrt{7} }{4}-\frac{3}{4} }{\frac{\sqrt{7} }{3}+\frac{3\sqrt{7} }{7} }=\frac{\frac{\sqrt{7}-3 }{4} }{\frac{7\sqrt{7} }{21} +\frac{9\sqrt{7} }{21} }=\frac{\frac{\sqrt{7}-3 }{4} }{\frac{16\sqrt{7} }{21} }=\frac{\sqrt{7}-3 }{4}*\frac{21}{16\sqrt{7} }=\frac{21\sqrt{7}-63 }{64\sqrt{7} }*\frac{\sqrt{7} }{\sqrt{7} }=\frac{-63\sqrt{7} +147}{448}[/tex]
Zad.3
[tex]2sin30+4(ctg30)^{2}-(tg30)^{2} = 2*\frac{1}{2}+4\sqrt{3}^{2}-(\frac{\sqrt{3} }{3})^{2}=1+4*3-\frac{3}{9}=1+12-\frac{1}{3}=12\frac{2}{3}[/tex]
Zad.4
[tex]cos\alpha =\frac{1}{2}[/tex] ⇒ [tex]\alpha =60[/tex]
[tex]tg\beta =1[/tex] ⇒ [tex]\beta=45[/tex]
γ - miara trzeciego kąta trójkąta ABC
γ + 45° + 60° = 180°
γ + 105° = 180°
γ = 180° - 105° = 75°
Odp. Miary kątów tego trójkąta to: 45°, 60°, 75°.