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°.

 

 

equilateral triangle with three sides labelled as having length '2'

 

Then let's drop the perpendicular, splitting the triangle into two 30-60-90 triangles:

We'll work with that 60° angle, as labelled.

 

 

triangle with altitude marked, 60-degree angle marked, half of base being 1

 

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

 

triangle placed into first quadrant
 

  

 
Now let's throw out the triangle, keeping just the base, the height, and the angle itself, along with the
x- and y-values from the point on the terminal side of the angle which marks where the hypotenuse of the original 30-60-90 triangle had ended.

 

x- and y-axes marked, terminal side of 60-degree angle marked, height line marked

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 same-length 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.

 

circle with radius r, centered at origin, added

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 x-axis. 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 x-value: 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!

 

60-degree line drawn in second quadrant, with base having length -1

Does this continue to work in the other quadrants? You bet!

 

 

Draw the corresponding line in QIII, marking 240°, being 60° below the negative x-axis. You can see from the picture that cos(240°) ought to be –1/2.

What does your calculator say?

 

line drawn in third quadrant marking terminal side of 240-degree angle, with base -1

 

 

 

Now move to QIV and draw the line for 300°, being 60° below the positive x-axis. In QIV, x is again positive, so cos(300°) = x/r = 1/2 = 0.5.

What does your calculator say?

 

line drawn in fourth quadrant, marking terminal side of 300-degree angle, with base +1

Note that in each case, all that really mattered was the sixty-degree triangle in the quadrant where the angle terminated, and the x- and y-values 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 x-axis, and then work from there. This works for all angles....


Original URL: http://www.purplemath.com/modules/quadangs.htm

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