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# Find the exact values of the six trigonometric functions of θ if θ is in standard position and the terminal side of θ is in the specified quadrant and satisfies the given condition. III; parallel to the line 5y − 2x + 1 = 0

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.