Odpowiedź:
[tex]zad.1\\\\d_{1} =10 cm~~,d_{2} =14 cm,~~ d_{1} ,d_{2} -przekatne ~~rombu~\\\\a-bok~~rombu\\\\z~~tw.Pitagorasa~~ obliczam~~a\\\\a^{2} =7^{2} +5^{2} \\\\a^{2} =49+25\\\\a^{2} =74\\\\a=\sqrt{74} \\\\sin\alpha =\frac{5}{\sqrt{74} } =\frac{5}{\sqrt{74} } \cdot\frac{\sqrt{74} }{\sqrt{74} } =\frac{5\sqrt{74} }{74} \\\\cos\alpha =\frac{7}{\sqrt{74} } =\frac{7}{\sqrt{74} } \cdot\frac{\sqrt{74} }{\sqrt{74} } =\frac{7\sqrt{74} }{74}\\\\ctg\alpha =\frac{7}{5} \\\\tg\alpha =\frac{5}{7} \\\\[/tex]
[tex]sin\beta =\frac{7}{\sqrt{74} } =\frac{7}{\sqrt{74} } \cdot\frac{\sqrt{74} }{\sqrt{74} } =\frac{7\sqrt{74} }{74}\\\\cos\beta =\frac{5}{\sqrt{74} } =\frac{5}{\sqrt{74} } \cdot\frac{\sqrt{74} }{\sqrt{74} } =\frac{5\sqrt{74} }{74}\\\\tg\beta =\frac{7}{5} \\\\ctg\beta =\frac{5}{7}[/tex]
[tex]zad.2\\a)\\sin45=\frac{\sqrt{2} }{2},~~tg30=\frac{\sqrt{3} }{3} ,~~cos60=\frac{1}{2}\\\\sin^{2}45-tg^{2} 30+cos^{2}60=(\frac{\sqrt{2} }{2})^{2} -(\frac{\sqrt{3} }{3})^{2} +(\frac{1}{2})^{2} =\frac{2}{4}-\frac{3}{9} +\frac{1}{4}= \frac{3}{4} -\frac{1}{3}=\frac{9}{12} -\frac{4}{12} =\frac{5}{12} \\\\b)\\sin60=\frac{\sqrt{2} }{2} ,~~tg45=1,~~cos30=\frac{\sqrt{3} }{2} \\\\sin^{2} 60-tg^{2}45+cos^{2} 30=(\frac{\sqrt{2} }{2})^{2} -1^{2}+(\frac{\sqrt{3} }{2})^{2} =\frac{2}{4} -1+\frac{3}{4}=\frac{1}{4}[/tex]
Szczegółowe wyjaśnienie:
