So the equation of the line is [tex]5y -2x + 1 = 0[/tex].
Let's put that in standard form (by solving for y):
[tex]5y -2x + 1 = 0[/tex]
[tex]5y = 2x - 1[/tex] ~ we added 2x and subtracted 1 from both sides
[tex]y = \frac{2}{5}x - \frac{1}{5}[/tex] ~ divide both sides by 5
So now we have it in the standard form which is [tex] y = mx + b [/tex].
The [tex] m [/tex] gives us the slope. Recall that slope is rise over run.
So the slope is [tex] \frac{2}{5} [/tex], 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:
[tex]hyp = \sqrt{2^2 + 5^2} = \sqrt{4+25} = \sqrt{29} [/tex]
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: [tex] \sqrt{29} [/tex]
Remember SOHCAHTOA. In order from that mnemonic:
sin(θ) = opposite over hypotenuse = [tex] \frac{2}{\sqrt{29}}[/tex]
cos(θ) = adjacent over hypotenuse = [tex] \frac{5}{\sqrt{29}} [/tex]
tan(θ) = opposite over adjacent = [tex] \frac{2}{5} [/tex]
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) = [tex] \frac{\sqrt{29}}{2} [/tex]
sec(θ) = [tex] \frac{\sqrt{29}}{5} [/tex]
cot(θ) = [tex] \frac{5}{2} [/tex]
We don't have to 'solve' (approximate) that square root in any of the answers since the problem asked for the exact solution.
