Finding trig values given a line

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The problem involves finding the trigonometric values sin θ, cos θ, and tan θ for an angle θ whose terminal side lies on the line 3x + 7y = 0 in quadrant III. The instructor suggested treating the triangle as if it were in quadrant II, leading to side lengths of 3 and 7. Using the Pythagorean theorem, the hypotenuse is calculated as √58. The resulting trigonometric values are sin θ = -3√58/58, cos θ = -7√58/58, and tan θ = 3/7, which appear to be correct. The question's complexity is acknowledged, but the calculations align with the given conditions.
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Homework Statement



The terminal side of angle θ in standard position lies on the given line in the given quadrant. Find sin θ, cos θ, and tan θ.

Homework Equations



3x + 7y = 0; quadrant III

The Attempt at a Solution



The line does not go through the third quadrant. The course instructor commented on this problem. I think he said to drop a line in the second quadrant instead, then to treat the sides of the triangle as if the triangle was in the third quadrant. I'm not sure what side lengths to use though and the relationship between the line, second, and third quadrant is not yet clear to me.
 
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If I solve for y, I get -3x/7 = y, giving me side lengths 3 and 7 in quadrant II. using Pythagorean theorem, I get sqrt (58) for the hypotenuse.


for quadrant 3:

sin θ = -3sqrt(58)/58
cos θ = -7sqrt(58)/58
tan θ = 3/7

Is this correct?
 
They look correct to me.
It's a weird question though.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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