Question

question 16 (1 point)
the point (-4, -6) is on the terminal arm of dc. which is the set of exact primary trigonometric ratios for the angle?
o a)
(sin c=-\frac{2}{sqrt{13}}, cos c=-\frac{3}{sqrt{13}}),
(\tan c=\frac{3}{2})
o b)
(sin c=-\frac{3}{sqrt{13}}, cos c=-\frac{2}{sqrt{13}}),
(\tan c=\frac{2}{3})
o c)
(sin c=\frac{3}{sqrt{13}}, cos c=\frac{2}{sqrt{13}}),
(\tan c=-\frac{2}{3})
o d)
(sin c=-\frac{3}{sqrt{13}}, cos c=-\frac{2}{sqrt{13}}),
(\tan c=\frac{3}{2})

Explanation:

Step1: Calculate the radius \( r \)

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

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 \(\tan C=\frac{y}{x}\)

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

Answer:

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