question 17 (1 point)\nthe point $(-2,5)$ is on the terminal arm of $db$. which is the set of exact primary…

question 17 (1 point)\nthe point $(-2,5)$ is on the terminal arm of $db$. which is the set of exact primary trigonometric ratios for the angle?\n\na)\n$sin b=-\frac{sqrt{29}}{2}, cos b=\frac{sqrt{29}}{5}$,\n$\tan b=-\frac{2}{5}$\n\nb)\n$sin b=-\frac{2}{sqrt{29}}, cos b=\frac{5}{sqrt{29}}$,\n$\tan b=-\frac{5}{2}$\n\nc)\n$sin b=\frac{5}{sqrt{29}}, cos b=-\frac{2}{sqrt{29}}$,\n$\tan b=-\frac{5}{2}$\n\nd)\n$sin b=\frac{sqrt{29}}{5}, cos b=-\frac{sqrt{29}}{2}$,\n$\tan b=-\frac{2}{5}$

question 17 (1 point)\nthe point $(-2,5)$ is on the terminal arm of $db$. which is the set of exact primary trigonometric ratios for the angle?\n\na)\n$sin b=-\frac{sqrt{29}}{2}, cos b=\frac{sqrt{29}}{5}$,\n$\tan b=-\frac{2}{5}$\n\nb)\n$sin b=-\frac{2}{sqrt{29}}, cos b=\frac{5}{sqrt{29}}$,\n$\tan b=-\frac{5}{2}$\n\nc)\n$sin b=\frac{5}{sqrt{29}}, cos b=-\frac{2}{sqrt{29}}$,\n$\tan b=-\frac{5}{2}$\n\nd)\n$sin b=\frac{sqrt{29}}{5}, cos b=-\frac{sqrt{29}}{2}$,\n$\tan b=-\frac{2}{5}$

Answer

Explanation:

Step1: Calculate the radius ( r )

For a point ((x,y)=(-2,5)) on the terminal arm of an angle, use the formula ( r=\sqrt{x^{2}+y^{2}} ). [ r = \sqrt{(-2)^{2}+5^{2}}=\sqrt{4 + 25}=\sqrt{29} ]

Step2: Calculate (\sin B), (\cos B) and (\tan B)

The trigonometric ratios are defined as (\sin B=\frac{y}{r}), (\cos B=\frac{x}{r}) and (\tan B=\frac{y}{x})

  • (\sin B=\frac{5}{\sqrt{29}}) (since (y = 5) and (r=\sqrt{29}))
  • (\cos B=\frac{-2}{\sqrt{29}}) (since (x=-2) and (r = \sqrt{29}))
  • (\tan B=\frac{5}{-2}=-\frac{5}{2}) (since (y = 5) and (x=-2))

Answer:

C. (\sin B=\frac{5}{\sqrt{29}},\cos B =-\frac{2}{\sqrt{29}},\tan B=-\frac{5}{2})