Question

question 15 (1 point) saved
the point (-5, -12) is on the terminal arm of dc.which is the set of exact reciprocal trigonometric ratios for the angle?
o a)
csc c = - 5/13, sin c = - 12/13,
cot c = 5/12
o b)
csc c = - 5/12, sec c = - 5/13,
cot c = 5/12
o c)
csc c = - 12/5, cos c = - 5/12,
cot c = 5/12
o d)
csc c = - 12/5, sec c = - 13/5,
cot c = 5/12

Explanation:

Step1: Calculate the radius \(r\)

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

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 \(x=-5\), \(y = - 12\), \(r = 13\) into the formulas:

• \(\csc C=\frac{r}{y}=\frac{13}{-12}=-\frac{13}{12}\) (This formula is wrong in options a, b, c. Let's check \(\csc C\) and \(\sec C\) again)
• \(\sec C=\frac{r}{x}=\frac{13}{-5}=-\frac{13}{5}\)
• \(\cot C=\frac{x}{y}=\frac{-5}{-12}=\frac{5}{12}\)

Answer:

d) \(\csc C=-\frac{13}{12}\), \(\sec C =-\frac{13}{5}\), \(\cot C=\frac{5}{12}\) (Note: There is a typo in the problem - \(\csc C\) formula in options. If we assume the problem means \(\sin C=\frac{y}{r}=-\frac{12}{13}\), \(\csc C=\frac{1}{\sin C}=-\frac{13}{12}\), \(\cos C=\frac{x}{r}=-\frac{5}{13}\), \(\sec C=\frac{1}{\cos C}=-\frac{13}{5}\), \(\cot C=\frac{\cos C}{\sin C}=\frac{5}{12}\). Among the given options, only option d has \(\csc C =-\frac{12}{5}\) (wrong value, but if we consider reciprocal of \(\sin C =-\frac{5}{13}\) is wrong in a, b, c. And for \(\sec C\), in option d \(\sec C=-\frac{13}{5}\) (since \(r = 13,x=-5\)) and \(\cot C=\frac{5}{12}\) (correct))