the point (-4, -6) is on the terminal arm of ∠c. which is the set of exact primary trigonometric ratios for…

the point (-4, -6) is on the terminal arm of ∠c. which is the set of exact primary trigonometric ratios for the angle?\na) sin c = -\\frac{3}{\\sqrt{13}}, cos c = -\\frac{2}{\\sqrt{13}}, tan c = \\frac{2}{3}\nb) sin c = -\\frac{3}{\\sqrt{13}}, cos c = -\\frac{2}{\\sqrt{13}}, tan c = \\frac{3}{2}\nc) sin c = -\\frac{2}{\\sqrt{13}}, cos c = -\\frac{3}{\\sqrt{13}}, tan c = \\frac{3}{2}\nd) sin c = \\frac{3}{\\sqrt{13}}, cos c = \\frac{2}{\\sqrt{13}}, tan c = -\\frac{2}{3}

the point (-4, -6) is on the terminal arm of ∠c. which is the set of exact primary trigonometric ratios for the angle?\na) sin c = -\\frac{3}{\\sqrt{13}}, cos c = -\\frac{2}{\\sqrt{13}}, tan c = \\frac{2}{3}\nb) sin c = -\\frac{3}{\\sqrt{13}}, cos c = -\\frac{2}{\\sqrt{13}}, tan c = \\frac{3}{2}\nc) sin c = -\\frac{2}{\\sqrt{13}}, cos c = -\\frac{3}{\\sqrt{13}}, tan c = \\frac{3}{2}\nd) sin c = \\frac{3}{\\sqrt{13}}, cos c = \\frac{2}{\\sqrt{13}}, tan c = -\\frac{2}{3}

Answer

Explanation:

Step1: Calculate the radius ( r )

Given the point ((x,y)=(-4,-6)), use the formula ( r=\sqrt{x^{2}+y^{2}} ). [ \begin{align*} r&=\sqrt{(-4)^{2}+(-6)^{2}}\ &=\sqrt{16 + 36}\ &=\sqrt{52}\ &=2\sqrt{13} \end{align*} ]

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

Use the trigonometric ratios: (\sin C=\frac{y}{r}), (\cos C=\frac{x}{r}), (\tan C=\frac{y}{x})

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

Answer:

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