Question 1155732
<pre>
He may be a little beyond what you're studying.  I think you're just studying
right-triangle trig, not circular functions, and haven't gotten to putting them
on a graph yet, so I'll do them assuming you are just doing right triangle trig.

Since sine is opposite/hypotenuse, draw a right triangle with the opposite side
as 1 and the hypotenuse as 5 and the angle on the left θ:

{{{drawing(400,4000/29,-0.5,5.3,-.5,1.5,
locate(5,.6,1), locate(2.5,.8,5), locate(1.3,.25,theta),
line(0,0,sqrt(24),0),line(sqrt(24),0,sqrt(24),1), line(sqrt(24),1,0,0) )}}}

Then use the Pythagorean theorem to find the adjacent side to θ:

{{{matrix(7,3,

c^2,""="",a^2+b^2,
5^2,""="",a^2+1^2,
25,""="",a^2+1,
24,""="",a^2,
sqrt(24),""="",a,
sqrt(4*6),""="",a,
2sqrt(6),""="",a)}}}

{{{drawing(400,4000/29,-0.5,5.3,-.5,1.5, locate(2.3,0,2sqrt(6)),
locate(5,.6,1), locate(2.5,.8,5), locate(1.3,.25,theta),
line(0,0,sqrt(24),0),line(sqrt(24),0,sqrt(24),1), line(sqrt(24),1,0,0) )}}}

{{{matrix(6,5,

sin(theta),""="",opposite/hypotenuse,""="",1/5,
cos(theta),""="",adjacent/hypotenuse,""="",2sqrt(6)/5,
tan(theta),""="",opposite/adjacent,""="",1/(2sqrt(6))=sqrt(6)/12,
sec(theta),""="",hypotenuse/adjacent,""="",2sqrt(6)/5,
csc(theta),""="",hypotenuse/opposite,""="",5/1=5,
cot(theta),""="",adjacent/opposite=1/5,""="",2sqrt(6)/1=2sqrt(6) )}}}

Edwn</pre>