Question

Please help ASAP! The terminal side of an angle θ in standard position passes through the point (2,-5). calculate the values of the six trigonometric functions for angle θ.

Answer 1

Answer:

$sin(x)=\frac{opp}{hyp}=\frac{5}{\sqrt{21}}=\frac{5\sqrt{21}}{21}\\cos(x)=\frac{adj}{hyp}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}\\tan(x)=\frac{opp}{adj}=\frac{5}{2}\\csc(x)=\frac{hyp}{opp}=\frac{\sqrt{21}}{5}\\sec(x)=\frac{hyp}{adj}=\frac{\sqrt{21}}{2}\\cot(x)=\frac{adj}{opp}=\frac{2}{5}$

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:

$sin(x)=\frac{opp}{hyp}=\frac{5}{\sqrt{21}}=\frac{5\sqrt{21}}{21}\\cos(x)=\frac{adj}{hyp}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}\\tan(x)=\frac{opp}{adj}=\frac{5}{2}\\csc(x)=\frac{hyp}{opp}=\frac{\sqrt{21}}{5}\\sec(x)=\frac{hyp}{adj}=\frac{\sqrt{21}}{2}\\cot(x)=\frac{adj}{opp}=\frac{2}{5}$

(Suppose that x is theta as I can't type it)

By some basic understanding that's all that I can do.