Question

The line with equation −a+3b=0 coincides with the terminal side of an angle θ in standard position and sinθ<0 .

Answer 1

If $a$ is the variable of the horizontal axis, then you can solve for $b$ to get the equation of the line in slope-intercept form in the $a,b$ plane:

$-a+3b=0\implies b=\dfrac a3$

i.e. a line with slope $\dfrac13$ through the origin, which means it is contained in the first and third quadrants. Since the terminal side of $\theta$ has a negative sine, the angle must lie in the third quadrant.

Because the slope of the line is $\dfrac13$, you can choose any length along the line to make up the hypotenuse of a right triangle with reference angle $\theta$. Any such right triangle will have $\tan\theta=\dfrac13$, regardless of whether the angle is the first or third quadrant. But since $\theta$ is known to lie in the third quadrant, and so $\sin\theta$ and $\cos\theta$ are both negative, you have

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac13\implies \dfrac{\sin\theta}{\cos\theta}=\dfrac{-1}{-3}\implies \cos\theta=-3$

Answer 2

Answer:

-310√10

Step-by-step explanation: