So the equation of the line is \(5y -2x + 1 = 0\).
Let's put that in standard form (by solving for y):
\(5y -2x + 1 = 0\)
\(5y = 2x - 1\) ~ we added 2x and subtracted 1 from both sides
\(y = \frac{2}{5}x - \frac{1}{5}\) ~ divide both sides by 5
So now we have it in the standard form which is \(y = mx + b\).
The \(m\) gives us the slope. Recall that slope is rise over run.
So the slope is \(\frac{2}{5}\), which means that you could draw a triangle with a rise of 2 and a run of 5.
That's almost enough information to find the trig values, but we still need to know the hypotenuse. Well, thanks to the magic of the Pythagorean Theorem:
\(hyp = \sqrt{2^2 + 5^2} = \sqrt{4+25} = \sqrt{29}\)
Alright, now that we know all three sides, we can find the values.
The "rise" (vertical) or opposite: 2
The "run" (horozontial) or adjacent: 5
The hypotenuse: \(\sqrt{29}\)
Remember SOHCAHTOA. In order from that mnemonic:
sin(θ) = opposite over hypotenuse = \(\frac{2}{\sqrt{29}}\)
cos(θ) = adjacent over hypotenuse = \(\frac{5}{\sqrt{29}}\)
tan(θ) = opposite over adjacent = \(\frac{2}{5}\)
Now the others (cosecant, secant, and cotangent) are just the reciprocals (upside-downs) of the others. Remember that cosecant comes first (since sine comes first in the "usuals".
csc(θ) = \(\frac{\sqrt{29}}{2}\)
sec(θ) = \(\frac{\sqrt{29}}{5}\)
cot(θ) = \(\frac{5}{2}\)
We don't have to 'solve' (approximate) that square root in any of the answers since the problem asked for the exact solution.
