Question

question 17 (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)
$sin b=-\frac{sqrt{29}}{2}, cos b=\frac{sqrt{29}}{5}$,
$\tan b=-\frac{2}{5}$

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

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

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

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 \(\tan B\)

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