جواب فعالیت صفحه 42 درس 2 ریاضی دهم (مثلثات)

الف

\(\begin{array}{l}x = \sqrt {{4^2} + {3^2}} = \sqrt {25} = 5\\\\\sin \theta = \frac{{BC}}{{AC}} = \frac{3}{5}\\\\\sin \alpha = \,\frac{4}{5}\\\\\cos \theta = \,\frac{4}{5}\\\\\cos \alpha = \frac{{BC}}{{AC}} = \frac{{\;3\,}}{5}\\\\\tan \theta = \frac{{BC}}{{AB}} = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\frac{3}{5}}}{{\frac{4}{5}}} = \frac{3}{4}\\\\\tan \alpha = \frac{{AB}}{{BC}} = \frac{{\frac{4}{5}}}{{\frac{3}{5}}} = \frac{4}{3}\\\\\cot \theta = \frac{1}{{\tan \theta }} = \frac{{\cos \theta }}{{\sin \theta }} = \frac{4}{3}.\\\\\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{3}{4}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{{{\sin }^2}\:\theta + {{\cos }^2}\:\theta = {{\left( {\frac{3}{5}} \right)}^2} + {{\left( {\frac{4}{5}} \right)}^2}}\\{ = \frac{9}{{25}} + \frac{{16}}{{25}} = \frac{{25}}{{25}} = 1}\\{}\\{{{\sin }^2}\:\alpha + {{\cos }^2}\:\alpha = {{\left( {\frac{4}{5}} \right)}^2} + {{\left( {\frac{3}{5}} \right)}^2}}\\{ = \frac{{16}}{{25}} + \frac{9}{{25}} = \frac{{25}}{{25}} = 1}\end{array}\)

پ

\(\begin{array}{l}{\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = {\sin ^2}\theta + {\cos ^2}\theta \\\\ = {\left( {\frac{{BC}}{{AC}}} \right)^2} + {\left( {\frac{{AB}}{{AC}}} \right)^2} = \frac{{B{C^2} + A{B^2}}}{{A{C^2}}}\\\\ = \frac{{A{C^2}}}{{A{C^2}}} = 1\end{array}\)

ت

\(\begin{array}{l}{\left( {\sin \:\alpha } \right)^2} + {\left( {\cos \:\alpha } \right)^2} = {\sin ^2}\:\alpha + {\cos ^2}\:\alpha \\\\ = {\left( {\frac{{AB}}{{AC}}} \right)^2} + {\left( {\frac{{BC}}{{AC}}} \right)^2}\\\\ = \frac{{A{B^2} + B{C^2}}}{{A{C^2}}} = \frac{{A{C^2}}}{{A{C^2}}} = 1\end{array}\)