Question

question 1 (1 point)
the point (-2, 5) is on the terminal arm of db.which is the set of exact primary trigonometric ratios for the angle?

a)
\\( \sin b=-\frac{2}{\sqrt{29}}, \cos b=\frac{5}{\sqrt{29}}, \\)
\\( \tan b=-\frac{5}{2} \\)

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

c)
\\( \sin b=\frac{5}{\sqrt{29}}, \cos b=-\frac{2}{\sqrt{29}}, \\)
\\( \tan b=-\frac{5}{2} \\)

d)
\\( \sin b=-\frac{\sqrt{29}}{2}, \cos b=\frac{\sqrt{29}}{5}, \\)
\\( \tan b=-\frac{2}{5} \\)

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 \(\tan B\)

Use the definitions \(\sin B=\frac{y}{r}\), \(\cos B=\frac{x}{r}\) and \(\tan B=\frac{y}{x}\).

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

Answer:

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