Quadrants
and Angles (page
1 of 3)
Sections: Introduction, Worked
Examples (and Sign Chart),
More Examples
You've worked with trig ratios in a geometrical
context: the context of right triangles. Now we'll move those ratios into
an algebraic context, and then we'll dispense with the triangles.
Let's start with an equilateral
triangle with sides of length 2:
Each angle of this triangle measures
60°.



Then let's drop the perpendicular,
splitting the triangle into two 306090
triangles:
We'll work with that 60° angle,
as labelled.



Now for the algebra: We'll
place the triangle on the x,yplane,
with the "adjacent" side on the xaxis,
the "opposite" side parallel to the yaxis,
and the angle we're working with, that 60°
angle, at the origin:



Now let's throw out the
triangle, keeping just the base, the height, and the angle itself,
along with the x
and yvalues
from the point on the terminal side of the angle which marks where
the hypotenuse of the original 306090
triangle had ended.



Since the length of the base is the same
as the value of "adjacent", we can call this just "x".
Similarly, the "opposite" is now just "y".
The Pythagorean Theorem gives us
the value of the hypotenuse, which we will now call "r".
Why? Because we can also look at this distance along the terminal
side of the angle as being a radius line of a circle.
Any right triangle with the samelength
hypotenuse will have its far end lying on this circle.
We already know that cos(60°)
= 1/2. Now you can read this
value from the drawing in the quadrants. The cosine ratio had been
"adjacent over hypotenuse"; now it is "x
over r".
But the value is still 1/2.



So we've set up the angle in the first
quadrant, thrown away the triangle, and still been able to find the trig
ratio. Can we build on this? Yes!
Let's swing over to the
second quadrant, drawing a similar line for 120°.,
being 60°
above the negative xaxis.
Since we're working in the same circle, we get the same value for
r,
but since we're in QII,
we get a negative xvalue:
x = –1
instead of +1.
Now grab your calculator, make sure
it's in "degree" mode, and plug in "cos(120°)".
Did you get "–0.5"?
That's the same as –1/2,
right?
Now look again at the graph: x/r
= (–1)/2 = –1/2 = –0.5, just
like your calculator said!



Does this continue to work in the other
quadrants? You bet!
Draw the corresponding line in QIII,
marking 240°,
being 60°
below the negative xaxis.
You can see from the picture that cos(240°)
ought to be –1/2.
What does your calculator say?



Now move to QIV
and draw the line for 300°,
being 60°
below the positive xaxis.
In QIV,
x
is again positive, so cos(300°)
= x/r = 1/2 = 0.5.
What does your calculator say?



Note that in each case, all that really
mattered was the sixtydegree triangle in the quadrant where the
angle terminated, and the x
and yvalues
in that quadrant. The fact that you can't draw a right triangle
with a base angle of 300°
was irrelevant. To find the trig ratios, you needed only to draw the terminal
side of the angle in whatever quadrant it happened to land, construct
a right triangle by "dropping a perpendicular" to the xaxis,
and then work from there. This works for all angles....
Original URL: http://www.purplemath.com/modules/quadangs.htm Copyright 2009 Elizabeth Stapel; All Rights Reserved. Terms of Use:
http://www.purplemath.com/terms.htm
