Question

the point (-5,-12) is on the terminal arm of \\( \angle c \\). which is the set of exact reciprocal trigonometric ratios for the angle?
a)\\( \csc c = - \frac { 12 } { 5 }, \cos c = - \frac { 5 } { 12 }, \cot c = \frac { 5 } { 12 } \\)
b)\\( \csc c = - \frac { 12 } { 5 }, \sec c = - \frac { 13 } { 5 }, \cot c = \frac { 5 } { 12 } \\)
c)\\( \csc c = - \frac { 5 } { 12 }, \sec c = - \frac { 5 } { 13 }, \cot c = \frac { 5 } { 12 } \\)
d)\\( \csc c = - \frac { 5 } { 13 }, \sec c = - \frac { 12 } { 13 }, \cot c = \frac { 5 } { 12 } \\)

Explanation:

Step1: Calculate the radius \(r\)

For a point \((x,y)=(-5,-12)\) on the terminal arm of an angle, use the formula \(r = \sqrt{x^{2}+y^{2}}\).
So, \(r=\sqrt{(-5)^{2}+(-12)^{2}}=\sqrt{25 + 144}=\sqrt{169}=13\).

Step2: Recall the reciprocal trigonometric ratios

The reciprocal trigonometric ratios are:

• \(\csc C=\frac{r}{y}\), \(\sec C=\frac{r}{x}\), \(\cot C=\frac{x}{y}\)

Substitute \(x=-5\), \(y = - 12\), \(r = 13\)

• \(\csc C=\frac{13}{-12}=-\frac{13}{12}\) (This is incorrect in the options, but let's check the other ratios)
• \(\sec C=\frac{13}{-5}=-\frac{13}{5}\)
• \(\cot C=\frac{-5}{-12}=\frac{5}{12}\)

Answer:

B. \(\csc C=-\frac{13}{12},\sec C =-\frac{13}{5},\cot C=\frac{5}{12}\)