Odpowiedź:
Ćw 3.
a)
[tex]sin\alpha =\frac{5}{13} \\ cos\alpha =\frac{12}{13} \\ tg\alpha =\frac{5}{12} \\ \\ \frac{sin\alpha }{cos\alpha} =tg\alpha \\ \frac{\frac{5}{13} }{\frac{12}{13} } =\frac{5}{12} \\ \frac{5}{13} *\frac{13}{12} =\frac{5}{12} \\ \frac{5}{12} =\frac{5}{12} \\ L=P[/tex]
b)
[tex]sin\alpha =\frac{1}{\sqrt{10} } *\frac{\sqrt{10} }{\sqrt{10} } =\frac{\sqrt{10} }{10} \\ cos\alpha =\frac{3}{\sqrt{10} } *\frac{\sqrt{10} }{\sqrt{10} } =\frac{3\sqrt{10} }{10} \\ tg\alpha =\frac{1}{3} \\ \\ \frac{sin\alpha }{cos\alpha} =tg\alpha \\ \frac{\frac{\sqrt{10} }{10} }{\frac{3\sqrt{10} }{10}} =\frac{1}{3} \\ \frac{\sqrt{10} }{10}*\frac{10}{3\sqrt{10} } =\frac{1}{3} \\ \frac{1}{3} =\frac{1}{3} \\ L=P[/tex]
Ćw 4.
a)
Korzystamy z tabelki wartości trygonometrycznych:
[tex]sin45=\frac{\sqrt{2} }{2} \\ cos45=\frac{\sqrt{2} }{2}\\ tg45=1[/tex]
[tex]\frac{sin45}{cos45} =tg45\\ \frac{\frac{\sqrt{2} }{2}}{\frac{\sqrt{2} }{2}} =1\\ 1=1\\ L=P[/tex]
b)
Korzystamy z tabelki wartości trygonometrycznych:
[tex]sin60=\frac{\sqrt{3} }{2} \\ cos60=\frac{1}{2} \\ tg60=\sqrt{3} \\ \\ \frac{sin60}{cos60} =tg60\\ \frac{\frac{\sqrt{3} }{2} }{\frac{1}{2}} =\sqrt{3} \\ \frac{\sqrt{3} }{2}*2=\sqrt{3}\\ \sqrt{3}=\sqrt{3}\\ L=P[/tex]
