Question

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

Explanation:

Step1: Calculate the radius \( r \)

Given the point \((x,y)=(-4,-6)\), use the formula \( r=\sqrt{x^{2}+y^{2}} \).

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