question 21 (1 point)\nthe point (-4, -6) is on the terminal arm of ∠c. which is the set of exact primary\n…

question 21 (1 point)\nthe point (-4, -6) is on the terminal arm of ∠c. which is the set of exact primary\n trigonometric ratios for the angle?\n a) sin c = -3/√13, cos c = -2/√13, tan c = 3/2\n b) sin c = -3/2, cos c = -3/√13, tan c = 2/3\n c) sin c = -3/√13, cos c = -2/√13, tan c = 3/2\n d) sin c = √13/3, cos c = √13/2, tan c = 2/3\n

question 21 (1 point)\nthe point (-4, -6) is on the terminal arm of ∠c. which is the set of exact primary\n trigonometric ratios for the angle?\n a) sin c = -3/√13, cos c = -2/√13, tan c = 3/2\n b) sin c = -3/2, cos c = -3/√13, tan c = 2/3\n c) sin c = -3/√13, cos c = -2/√13, tan c = 3/2\n d) sin c = √13/3, cos c = √13/2, tan c = 2/3\n

Answer

Explanation:

Step1: Calculate the radius ( r )

For a point ((x,y)=(-4,-6)) in the coordinate - plane, 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)

The formula for (\sin C=\frac{y}{r}). Substitute (y=-6) and (r = 2\sqrt{13}) (\sin C=\frac{-6}{2\sqrt{13}}=-\frac{3}{\sqrt{13}}=-\frac{3\sqrt{13}}{13})

Step3: Calculate (\cos C)

The formula for (\cos C=\frac{x}{r}). Substitute (x = - 4) and (r=2\sqrt{13}) (\cos C=\frac{-4}{2\sqrt{13}}=-\frac{2}{\sqrt{13}}=-\frac{2\sqrt{13}}{13})

Step4: Calculate (\tan C)

The formula for (\tan C=\frac{y}{x}). Substitute (x=-4) and (y = - 6) (\tan C=\frac{-6}{-4}=\frac{3}{2})

Answer:

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