The terminal side of $\theta$ lies on the given line in the specified quadrant. Find the value of the given trigonometric function of $\theta$ by finding a point on the line.
$\begin{array} { l c c } \text { Line } & \text { Quadrant } & \text { Evaluate: } \ 11 x + 3 y = 0 & \text { II } & \csc \theta \end{array}$
A) $- \frac { \sqrt { 130 } } { 11 }$
B) $\frac { 11 } { \sqrt { 130 } }$
C) $\frac { \sqrt { 130 } } { 11 }$
D) $- \frac { 11 } { \sqrt { 130 } }$
E) $\frac { 11 } { 3 }$

Question

Quizplus · Accepted Answer

In the second quadrant, sine is positive. A point on the line $11x + 3y = 0$ can be $(-3, 11)$. The hypotenuse is $\sqrt{(-3)^2 + 11^2} = \sqrt{130}$. Thus, $\csc \theta = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{\sqrt{130}}{11}$.

The Terminal Side Of θ\thetaθ Lies on the Given Line in the Specified Quadrant

The Terminal Side Of $\theta$ Lies on the Given Line in the Specified Quadrant