Answer:
[tex]sin(x)=\frac{opp}{hyp}=\frac{5}{\sqrt{21}}=\frac{5\sqrt{21}}{21}\\cos(x)=\frac{adj}{hyp}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}\\tan(x)=\frac{opp}{adj}=\frac{5}{2}\\csc(x)=\frac{hyp}{opp}=\frac{\sqrt{21}}{5}\\sec(x)=\frac{hyp}{adj}=\frac{\sqrt{21}}{2}\\cot(x)=\frac{adj}{opp}=\frac{2}{5}[/tex]
Step-by-step explanation:
Start by drawing out the triangle on the graph:
(See picture)
(By the drawing you can judge me that I am a really bad artist)
Theta and the side lengths are labeled, the hypotenuse has a length of √21 by Pythagora's theorem. Now, time to put everything to the trigonometric functions:
[tex]sin(x)=\frac{opp}{hyp}=\frac{5}{\sqrt{21}}=\frac{5\sqrt{21}}{21}\\cos(x)=\frac{adj}{hyp}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}\\tan(x)=\frac{opp}{adj}=\frac{5}{2}\\csc(x)=\frac{hyp}{opp}=\frac{\sqrt{21}}{5}\\sec(x)=\frac{hyp}{adj}=\frac{\sqrt{21}}{2}\\cot(x)=\frac{adj}{opp}=\frac{2}{5}[/tex]
(Suppose that x is theta as I can't type it)
By some basic understanding that's all that I can do.
