Given that θ terminates in Quadrant III and tanθ 5/12= , find cscθ

[tex]\bf tan(\theta)=\cfrac{opposite}{adjacent} \qquad \qquad csc(\theta)=\cfrac{hypotenuse}{opposite} \\\\ -------------------------------\\\\ tan(\theta)=\cfrac{5}{12}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}\\\\ -------------------------------\\\\ \textit{let's use the pythagorean theorem to get the hypotenuse "c"} \\\\\\ c^2=a^2+b^2\implies c=\pm\sqrt{a^2+b^2}\implies c=\pm\sqrt{12^2+5^2} \\\\\\ \boxed{c=\pm 13}[/tex]

now, the square root gives us the +/- version, so.. which is it? well, the hypotenuse is just a radius unit, is never negative, just the radius unit, so just the absolute value of that or the version 13, so c = 13

now, that we know what the hypotenuse "c" is, well

[tex]\bf csc(\theta)=\cfrac{hypotenuse}{opposite}\implies csc(\theta)\cfrac{13}{5}[/tex]

dfrancolopera dfrancolopera · Answer 2 · 2019-09-18T20:40:48+02:00

Answer:

[tex]csc(\theta)=-\frac{13}{5}[/tex]

Step-by-step explanation:

We've been told that θ terminates in Quadrant III, which means that θ is the angle between the positive x-axis and a vector [tex]\vec{r}[/tex] with negative x and y coordinates.

Additionally, we know that [tex]tan(\theta)=\frac{5}{12}=\frac{y-coordinate}{x-coordinate}[/tex]

So, using this information we can conclude that:

x-coordinate= [tex]x=-12[/tex]

y-coordinate= [tex]y=-5[/tex]

(This make sense because [tex]tan(\theta)=\frac{-5}{-12}=\frac{5}{12}[/tex])

But, we still have to calculate [tex]csc(\theta)[/tex]

[tex]csc(\theta)=\frac{r}{y-coordinate}[/tex]

Where [tex]r[/tex] is the vector magnitude

[tex]r=\sqrt{x^2+y^2} =\sqrt{(-12)^2+(-5)^2}=13[/tex]

[tex]csc(\theta)=\frac{13}{(-5)}=-\frac{13}{5}[/tex]

So we've found the answer: [tex]csc(\theta)=-\frac{13}{5}[/tex]