Question

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

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