Question

question 6 (1 point)
the point (-2,5) is on the terminal arm of ∠b.which is the set of exact primary
trigonometric ratios for the angle?
a) sin b = -2/√29, cos b = 5/√29, tan b = -5/2
b) sin b = -√29/2, cos b = √29/5, tan b = -2/5
c) sin b = 5/√29, cos b = -2/√29, tan b = -5/2

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 \( x = - 2\), \( y = 5\), then \( r=\sqrt{(-2)^{2}+5^{2}}=\sqrt{4 + 25}=\sqrt{29} \).

Step2: Use trigonometric ratio definitions

• The sine function is defined as \( \sin B=\frac{y}{r} \). Substituting \( y = 5\) and \( r=\sqrt{29}\), we get \( \sin B=\frac{5}{\sqrt{29}}=\frac{5\sqrt{29}}{29} \).
• The cosine function is defined as \( \cos B=\frac{x}{r} \). Substituting \( x=-2\) and \( r = \sqrt{29}\), we get \( \cos B=\frac{-2}{\sqrt{29}}=-\frac{2\sqrt{29}}{29} \).
• The tangent function is defined as \( \tan B=\frac{y}{x} \). Substituting \( y = 5\) and \( x=-2\), we get \( \tan B=\frac{5}{-2}=-\frac{5}{2} \).

Answer:

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