question 32 (1 point)\nthe point (-4,-6) is on the terminal arm of dc. which is the set of exact primary\n…

question 32 (1 point)\nthe point (-4,-6) is on the terminal arm of dc. which is the set of exact primary\n trigonometric ratios for the angle?\n a) \n sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},\n tan c = \frac{3}{2}\n b) \n sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},\n tan c = \frac{3}{2}\n c) \n sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},\n tan c = \frac{3}{2}\n d) \n sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},\n tan c = \frac{3}{2}\n

question 32 (1 point)\nthe point (-4,-6) is on the terminal arm of dc. which is the set of exact primary\n trigonometric ratios for the angle?\n a) \n sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},\n tan c = \frac{3}{2}\n b) \n sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},\n tan c = \frac{3}{2}\n c) \n sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},\n tan c = \frac{3}{2}\n d) \n sin c = \frac{3}{sqrt{13}}, cos c = -\frac{2}{sqrt{13}},\n tan c = \frac{3}{2}\n

Answer

Explanation:

Step1: Calculate the hypotenuse ( r )

For a point ((x,y)=(-4,-6)) on the terminal arm of an angle, use the formula ( r=\sqrt{x^{2}+y^{2}} ). [ r=\sqrt{(-4)^{2}+(-6)^{2}}=\sqrt{16 + 36}=\sqrt{52}=2\sqrt{13} ]

Step2: Calculate (\sin C), (\cos C) and (\tan C)

The trigonometric ratios are defined as (\sin C=\frac{y}{r}), (\cos C=\frac{x}{r}) and (\tan C=\frac{y}{x}).

  • (\sin C=\frac{-6}{2\sqrt{13}}=\frac{-3}{\sqrt{13}}=\frac{-3\sqrt{13}}{13})
  • (\cos C=\frac{-4}{2\sqrt{13}}=\frac{-2}{\sqrt{13}}=\frac{-2\sqrt{13}}{13})
  • (\tan C=\frac{-6}{-4}=\frac{3}{2})

Answer:

d) (\sin C =-\frac{3\sqrt{13}}{13},\cos C =-\frac{2\sqrt{13}}{13},\tan C=\frac{3}{2})