SOLUTION: I am trying to evaluate Trig Functions determined by a point in Quadrant 1. When the problem gives a graph that goes along with it, that tells me where the (x,y) point is, along w

Question 733738: I am trying to evaluate Trig Functions determined by a point in Quadrant 1. When the problem gives a graph that goes along with it, that tells me where the (x,y) point is, along where the ray (that will become the hypotenuse of the triangle) is, then I can figure it out just fine. My problem is when it gives me a a question without a graph. Like when it is simply asking me to find the six trigonometric functions for theta when all I have is: P(3,4) and no graph to go along with it. I know the x and y is the adjacent and opposite of the triangle, respectively, but I do not know how to tell which is which unless I am looking at a graph about it. So my question is how do I find out whether the x or the y is the adjacent or opposite without a graph to look at?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
My problem is when it gives me a a question without a graph. Like when it is simply asking me to find the six trigonometric functions for theta when all I have is: P(3,4) and no graph to go along with it. I know the x and y is the adjacent and opposite of the triangle, respectively, but I do not know how to tell which is which unless I am looking at a graph about it. So my question is how do I find out whether the x or the y is the adjacent or opposite without a graph to look at?
----
plot the point (3,4)
Draw a line segment from (0,0) to (3,4)---That is the hypotenuse.
Draw a line to the x-axis at (3,0).
---
The base is 3 ; the opposite is 4 ; the hypotenuse is sqrt(3^2+4^2) = 5
-----------------------------
The angle is that made by the x axis and the hypotenuse.
---------------------------------
sin = y/r = 4/5
cos = x/r = 3/5
tan = y/x = 4/3
-----
csc = r/y = 5/4
sec = r/x = 5/3
cot = x/y = 3/4
==================
Cheers,
Stan H.
================
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