Today's soundtrack is *Rebecca St. James: If I Had One Chance To Tell You Something*. She is such an incredible singer; I've loved her music ever since I got her album *Transform* when I was in my teens.

Today, I'm continuing where I left off yesterday in learning about how we can measure and work with angles in all four quadrants.

When dealing with angles in the first quadrant, we often represent them as *θ* (theta); however, when we look at angles in quadrants 2, 3, or 4, we represent those angles as *α *(alpha). Another way to remember this is that "[f]or angles greater than 90°, α *represents the reference angle*" (Pearson's Pre-Calculus 11, p. 442).

A bit of absolute information on variables:

When dealing with angles in standard position, if *θ *is between 0° and 360°, we represent our sides in the following way:

*x*= adjacent*y*= opposite*r*= hypotenuse

We must consider whether our *x* and *y* coordinates of the terminal point are positive or negative when working out our ratios. *r, *being simply the length of the hypotenuse, will always be positive - we can't have a negative hypotenuse!

All angles in Quadrant 1 will always be positive, as both

*x*and*y*will be positive.sin

*θ*=*y*/*r*cos

*θ*=*x*/*r*tan

*θ = y*/*x***In Quadrant 2,*x*is negative, and*y*is positive, so any ratios that include*x*will be negative.sin

*α*=*y*/r**- cos***α***=***x*/*r*tan

*α = y*/*x***In Quadrant 3, both*x*and*y*are negative*,*so any angles that include either x or*y*will be negative; however, if they are together, the two negatives will make a positive.**- sin***α*=*y*/r**- cos***α***=***x*/*r*tan

*α = y*/*x*In Quadrant 4,

*x*is positive, and*y*is negative, so any ratios that include*y*will be negative.**- sin***θ*=*y*/*r*cos

*θ*=*x*/*r***- tan***θ = y*/*x*

A terminal point tells us right away what quadrant we are working with.

(+

*x*, +*y*) = Quadrant 1(-

*x*, +*y*) = Quadrant 2(-

*x,*-*y*) = Quadrant 3(+

*x*, -*y*) = Quadrant 4

That covers all the theory about angles in standard position in all quadrants! Now, there are three main things that we will be asked to do when dealing with these angles:

Find the ratios of an angle when we know a terminal point

Find the ratios within all quadrants when they have the same reference angle

Solve a triangle

Find information about ratios when we know about an angle

Let's walk through a quick "how-to" of each one.

**How to find the ratios of an angle when we know a terminal point**

If we know the *x *and *y *coordinates of our terminal point, we may be asked to find the ratios of *θ*, and then to determine the measure of *θ*. To do so, we will follow these steps:

Use the Pythagorean Theorem with our

*x*/*y*coordinates to determine the value of*r*Now that we know the lengths of all sides, write out the three ratios with those values included

Choose one positive ratio from step 2

Use the inverse ratio function to find the measure of the reference angle

Determine which quadrant our point is in by noting the positive/negative values of

*x*and*y*Determine the true angle

Quadrant 1:

**[reference angle] = angle**Quadrant 2:

**180 - [reference angle] = angle**Quadrant 3:

**[reference angle] - 180 = angle**Quadrant 4:

**360 - [reference angle] = angle**

**How to find the ratios of the same reference angle in each quadrant **

If we are given an angle with a terminal point , and we are told to find the ratios of all angles that have the same reference angle, we must do the following:

Use substitution and the ratio function on the calculator to determine the length of each side

For example, if we are told that the terminal arm's coordinates are

*r*cos*θ*and*r*sin*θ*, and we know that*θ*= 60° and that the length of the hypotenuse is 2, we will calculate "2 cos(60)" and "2 sin(60)". 2cos(60) = 1; divide both sides by 2 to isolate cos(60), giving us cos(60)=1/2. 2 sin(60) = 1.732; divided by 2, we get sin(60) = 1.732/2. Since cos = adj/hyp, and sin = opp/hyp, we now know the lengths of all three sides:*x*= 1,*y*= 1.732, and*r*= 2; we already knew that the reference angle we are looking at is 60°.

Find the angle in each quadrant that corresponds with the reference angle

Quadrant 1:

**[reference angle] = angle**Quadrant 2:

**180 - [reference angle] = angle**Quadrant 3:

**[reference angle] - 180 = angle**Quadrant 4:

**360 - [reference angle] = angle**

****Write out the ratios for each quadrant, substituting in known values

Quadrant 1

sin

*θ*=*y*/*r,*cos*θ*=*x*/*r,*tan*θ = y*/*x***Quadrant 2sin

*α*=*y*/r,**- cos****α****=****x****/**tan**r**,*α = y*/*x***Quadrant 3**- sin**,*α*=*y*/r**- cos***α***=***x*/*r**,*tan*α = y*/*x*Quadrant 4

**- sin***θ*=, cos*y*/*r**θ*=*x*/*r*,**- tan***θ = y*/*x*

Using the ratios found in step 3, write out the coordinates of the terminal point in each quadrant**Step 4 is very similar to Step 1. Just use the relevant angle measurement in the equations: for example, our fourth quadrant's angle is 300°, so to find the fourth quadrant's

*x*-coordinate, we need to find 2 cos(300), which is 1; again, divide the whole thing by*r*to determine that the*x*-coordinate is 1/2*.*We can calculate 2 sin(300) to determine that the*y-*coordinate of the terminal arm in Quadrant 4 is -1.732/2.

**How to solve a triangle **

When working with right triangles, if we know either the measure of one angle and one leg's length, or two legs' lengths, we can solve that triangle. Check out my right angle trigonometry flowchart!

**How to find information about ratios when we know about an angle**

If we are given an angle's ratio, we may be asked to find the other ratios and to find the other possible values of that angle.

Use the Pythagorean Theorem to extrapolate the third side's length

For example, if we are told that tan

*α*= 5/12, we already know*α*'s opposite and adjacent lengths, meaning that we just need to find the length of the hypotenuse (*r*).Based on whether the values we know are positive or negative, determine which other quadrants might have the same values for

*α*.In the above example, we know that tan

*α*is positive, meaning that either both*x*and*y*are positive, or both are negative. So tan*α*could be either in Quadrant 1 or in Quadrant 3.

That's all for today! What a doozy! Next time, I'll be working on the assignment portion of this lesson.