Finding the radian value of this angle which passes through a point

[Ace.]

Homework Statement

The terminal arm of an angle in standard position passes through (-7,8). Find the radian value of the angle in the interval [0,2∏], to the nearest hundredth.

Homework Equations

sinθ = \frac{y}{r}
cosθ = \frac{x}{r}
tanθ = \frac{y}{x}

The Attempt at a Solution

The terminal arm is in quadrant 2, and I found the side lengths to be -7,8, and \sqrt{113} (hypotenuse). When I tried to find the value of θ I get different answers for different ratios.

θ = sin^{-1}\frac{8}{\sqrt{113}}

= 0.85​

θ = cos^{-1}\frac{-7}{\sqrt{113}}

= 2.29​

θ = tan^{-1}\frac{8}{-7}

= -0.85​

The correct one is θ=2.29. Why is this correct and not the others?

 

[Chestermiller]

The point -7,8 is in the second quadrant. So θ is less than π and greater than π/2.

 

[Mark44]

Homework Statement

The terminal arm of an angle in standard position passes through (-7,8). Find the radian value of the angle in the interval [0,2∏], to the nearest hundredth.

Homework Equations

sinθ = \frac{y}{r}
cosθ = \frac{x}{r}
tanθ = \frac{y}{x}

The Attempt at a Solution

The terminal arm is in quadrant 2, and I found the side lengths to be -7,8, and \sqrt{113} (hypotenuse). When I tried to find the value of θ I get different answers for different ratios.

θ = sin^{-1}\frac{8}{\sqrt{113}}

= 0.85​

θ = cos^{-1}\frac{-7}{\sqrt{113}}

= 2.29​

θ = tan^{-1}\frac{8}{-7}

= -0.85​

The correct one is θ=2.29. Why is this correct and not the others?

Because, as you pointed out, the angle is in the 2nd quadrant. This means that θ is between $\pi/2$ and $\pi$ (roughly, between 1.5 and 3).

The sin^-1 function returns an angle between $-\pi/2$ and $\pi/2$, as does tan^-1. The cos^-1 function returns an angle between 0 $\pi$, which was the correct interval for your angle.