نویسنده : امیر انصاری

پاسخ

1. $$
   r=\sqrt{x^2+y^2}\\
   r=\sqrt{(-5)^2+12^2}=\sqrt{25+144}=\sqrt{169}=13\\
   \sin \theta = \frac{y}{r}=\frac{12}{13}\\
   \cos \theta=\frac{x}{r}=-\frac{5}{13}\\
   \tan \theta = \frac{y}{x}=-\frac{12}{5}
   $$
   
2. $$
   r=\sqrt{x^2+y^2}\\
   r=\sqrt{5^2+(-3)^2}=\sqrt{25+9}=\sqrt{34}\\
   \sin \theta = \frac{y}{r}=-\frac{3}{\sqrt{34}}\\
   \cos \theta = \frac{x}{r}=\frac{5}{\sqrt{34}}\\
   \tan \theta = \frac{y}{x} = -\frac{3}{5}
   $$
   
3. $$
   r=\sqrt{x^2+y^2}\\
   r=\sqrt{6^2+3^2}=\sqrt{36+9}=\sqrt{45}\\
   \sin \theta = \frac{y}{r} = \frac{3}{\sqrt{45}} = \frac{3}{3\sqrt{5}}=\frac{1}{\sqrt{5}}\\
   \cos \theta = \frac{x}{r} = \frac{6}{\sqrt{45}}=\frac{6}{3\sqrt{5}}=\frac{2}{\sqrt{5}}\\
   \tan \theta = \frac{y}{x} = \frac{3}{6}=\frac{1}{2}
   $$
   
4. $$
   r=\sqrt{x^2+y^2}\\
   r=\sqrt{(-24)^2+(-10)^2}=\sqrt{576+100}=\sqrt{676}=26\\
   \sin \theta = \frac{y}{r} = -\frac{10}{26}=-\frac{5}{13}\\
   \cos \theta = \frac{x}{r} = -\frac{24}{26}=-\frac{12}{13}\\
   \tan \theta = \frac{y}{x} = \frac{-10}{-24}=\frac{10}{24}=\frac{5}{12}
   $$