Question

the point (3, -5) is on the terminal arm of ∠d. which is the set of exact primary trigonometric ratios for the angle?
a) sin d = -\frac{\sqrt{34}}{3}, cos d = \frac{\sqrt{34}}{5}, tan d = -\frac{5}{3}
b) sin d = \frac{\sqrt{34}}{5}, cos d = -\frac{\sqrt{34}}{3}, tan d = -\frac{5}{3}
c) sin d = -\frac{\sqrt{34}}{5}, cos d = \frac{\sqrt{34}}{3}, tan d = -\frac{5}{3}
d) sin d = -\frac{\sqrt{34}}{5}, cos d = \frac{\sqrt{34}}{3}, tan d = -\frac{1}{3}

Explanation:

Step1: Calculate the radius \( r \)

For a point \((x,y)\) on the terminal side of an angle, \(r=\sqrt{x^{2}+y^{2}}\). Given \(x = 3\) and \(y=-5\), then \(r=\sqrt{3^{2}+(-5)^{2}}=\sqrt{9 + 25}=\sqrt{34}\).

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

The definitions of trigonometric functions are \(\sin D=\frac{y}{r}\), \(\cos D=\frac{x}{r}\) and \(\tan D=\frac{y}{x}\).
Since \(x = 3\), \(y=-5\) and \(r=\sqrt{34}\), we have:

• \(\sin D=\frac{y}{r}=\frac{-5}{\sqrt{34}}=-\frac{5\sqrt{34}}{34}\) (rationalize the denominator: \(\frac{-5}{\sqrt{34}}\times\frac{\sqrt{34}}{\sqrt{34}}=-\frac{5\sqrt{34}}{34}\))
• \(\cos D=\frac{x}{r}=\frac{3}{\sqrt{34}}=\frac{3\sqrt{34}}{34}\) (rationalize the denominator: \(\frac{3}{\sqrt{34}}\times\frac{\sqrt{34}}{\sqrt{34}}=\frac{3\sqrt{34}}{34}\))
• \(\tan D=\frac{y}{x}=\frac{-5}{3}\)

Answer:

b) \(\sin D =-\frac{5\sqrt{34}}{34},\cos D=\frac{3\sqrt{34}}{34},\tan D =-\frac{5}{3}\)