Draw Quadrant I of the XY axes: that is, just draw the top-right portion of how you would normally draw your XY-axes graph. That way you only see the positive x, positive y parts.

Now, on the horizontal axis, going right, draw a line. Where you stop, call this point C. Now draw a vertical line straight up from here, calling this point B where you stop. From the origin (call it point A), draw a ray connecting to point C and going on.

We say the length of AC is x, the length of CB is y, and then the length of AB is r. B has coordinats (x,y).

Now, the side length y is called the side opposite of angle A. This makes sense because y is on the opposite side of A. The side x is called the side adjacent of angle A. This makes sense because the two are right next to each other. Note that for adjacents, you DO NOT LOOK AT THE HYPOTENUSE, r. NEVER. ONLY CONSIDER THE TWO "legs."

Now, from there, we can define our trig functions, which consists of a table you should commit to memory.

sinA = y/r = side opposite / hypotenuse
cosA = x/r = side adjacent / hypotenuse
tanA = y/x = side opposite / side adjacednt
cscA = r/y = hypotenuse / side opposite
secA = r/x = hypotenuse / side adjacent
cotA = x/y = side adjacent / side opposite

Keep that drawing. We'll be using it later on.

Our next table you should commit to memory consists of cofunctions. Cofunctions are complementary, in that the sum of the angles 90^o. And it works like this--

sinA = cos(90^o - A)
cosA = sin(90^o - A)
cscA = sec(90^o - A)
secA = csc(90^o - A)
tanA = cot(90^o - A)
cotA = tan(90^o - A), for all A < 90^o.

How do we gather this? Well, look at our triangle. More specifically, let's look at sinA = y/r.

Now, look at angle B. Look at ITS side adjacent, which is y. We can say cosB = y/r. What else = y/r?

sinA!! Thus, we can say sinA = cosB. Now, we know there is a 90^o angle there called angle C based on our drawing. We know triangles have 180^o. So, we can eliminate that 90^o and get 180^o - 90^o = 90^o.

Thus, A + B = 90^o. Now, what say we subtract A from both sides? A - A + B = 90^o - A, or B = 90^o - A.

Look at that! We just saw how our cofunctions work together! In our table, instead of saying sinA = cosB, we simply replaced B with 90^o - A to get sinA = cos(90^o - A), and we did the SAME THING with every other function there. That's how cofunctions work!

Now, we need to simply analyze two particular types of triangles that come up VERY frequently (so why NOT give them special treatment?). They are dubbed the 30-60 triangle and the 45-45 triangle. We'll start with the 45-45 triangle.

The 45-45 triangle is simply an isosceles triangle with one angle = 90^o and the other two angles = 45^o.

Draw one and just specificy two angles as being 45^o with a 90^o angle there. We know based on how isosceles triangles work that the two sides opposite the two 45^o angles are congruent, or equal in measure, because their angles are congruent as well.

Now, specify any length to the two equal sides.  For simplicity, we'll say that these lengths = 1 each.  Using Pythagorean Theorem, we can, with a = b, say that c² = a² + b² = 1² + 1² = 1 + 1 = 2.  c = Sqrt(2).  We'll denote this "c" as "r", and we'll go ahead and denote the height a (or b) = y and the length b (or a) = x.

Now, taking the triangle to go down, right, up-left (the edges, I mean), this puts the right angle in the lower-left corner.  With x = length = 1 and y = height = 1 and r = hypotenuse = Sqrt(2), we can find the six trigonometric functions for the 45 degree angles.  Take the bottom right 45^o angle as our angle.  We can say that sin(45) = y / r = 1 / Sqrt(2) = Sqrt(2) / 2.  cos(45) = x / r = 1 / Sqrt(2) = Sqrt(2) / 2.  tan(45) = y / x = 1 / 1 = 1.  csc(45) = r / y = Sqrt(2) / 1 = Sqrt(2).  sec(45) = r / x = Sqrt(2) / 1 = Sqrt(2).  cot(45) = x / y = 1 / 1 = 1.

BAM.  We have all 6 functions for ALL angles of 45 degrees.  It does not matter what the lengths of the sides are:  sin(45) is ALWAYS Sqrt(2)/2.  ALWAYS.  Same with the others.

Now, let's look at that 30-60 triangle.  Rather than bother drawing a 30-60 triangle, let's draw an equilateral triangle (all lengths are the same, all angles = 60^o).  We'll let each length = 2.

Now, draw a vertical line to divide this sucker in half, and we'll look at the left half.  As we can easily see, the bottom has been split in half, so no longer is it = 2, but now it = 1.  We'll let this be our length x, and the height = y, and our hypotenuse = r.  Because we split the triangle in half, we also split that top angle in half.  Now, the top angle is 30^o, putting our 60^o down in the lower left corner, and our right angle is in the lower right.  Now, we need to find the length of this height.  By Pythagorean Theorem, we have r² = x² + y².  2² = 1² + y².  4 = 1 + y².  3 = y².  y = Sqrt(3).

Now that we have all three angles and all three sides, we can find the trigonometric functions.  We shall evaluate them at the 60^o angle.

sin(60) = y / r = Sqrt(3) / 2.  cos(60) = x / r = 1 / 2.  tan(60) = y / x = Sqrt(3) / 1 = Sqrt(3).  csc(60) = r / y = 2 / Sqrt(3) = 2Sqrt(3) / 3.  sec(60) = r / x = 2 / 1 = 2.  cot(60) = x / y = 1 / Sqrt(3) = Sqrt(3) / 3.

Now for a little trick.  If you analyze the 30^o angle, you see things are just the opposite.  That is, sin(30) = cos(60) = 1/2.

These are the most fundamental angles, 30, 45, and 60, and they make up something we'll look at later called the Unit Circle.  You better go ahead and start memorizing them because they are angles every aspiring math student should memorize (as you work with them later on in this course and in calculus if you do it, you use these angles so much you really can't help BUT memorize them, even accidentally).

To help, let's make a table!
       sine           cosine       tangent       cotangent   secant   cosecant
30   1/2            Sqrt(3)/2   Sqrt(3)/3       Sqrt(3)      2Sqrt(3)/3       2
45    Sqrt(2)/2   Sqrt(2)/2   1                   1               Sqrt(2)         Sqrt(2)
60   Sqrt(3)/2      1/2          Sqrt(3)          Sqrt(3)/3      2               2Sqrt(3)/3

Now for some problems!

1)  Given a right triangle with x = 20, y = 21, and r = 29, find sinA, cosA, and tanA where angle A is the angle opposite y.

sinA = y / r = 21/29.  cosA = x / r = 20/29.  tanA = y / x = 21/20.

2)  Write cot(73^o), cos(a + 20^o), and tan(25.4^o) in terms of their cofunctions.

Well, we see for the first we have cotA.  cotA = tan(90 - A), so this would be tan(90 - 73) = tan(17^o).
For the second one, it looks ugly with a variable, but don't worry about it.  Just do this normally.  cosA = sin(90 - A) = sin(90 - a + 20) = sin(70^o - a).  The last one is just tanA = cot(90 - A) = cot(90 - 25.4) = cot(64.6^o).

3)  Find one solution for cot(5B + 2^o) = tan(2B + 4^o).

Ew.  Well, just use your cofunction.  We know that cotA = tan(90 - A), so we can rewrite the cotangent part as simply tan(90 - A) = tan(90 - 5B - 2^o) = tan(88^o - 5B).  Putting that = tan(2B + 4^o), the tangents go away, leaving us with 88 - 5B = 2B + 4.  Now it's stupid simple.  84 - 5B = 2B.  84 = 7B.  B = 12.

The first thing we're going to talk about is the reciprocal identity.

Remember how in the last section we said sinθ = y/r, and then we said cscθ = r/y?  Look at that.  y/r, r/y. Why, that's a reciprocal!  THAT'S NO COINCIDENCE!

Let's list 'em!

sin(θ) = 1/csc(θ)
cos(θ) = 1/sec(θ)
tan(θ) = 1/cot(θ)
csc(θ) = 1/sin(θ)
sec(θ) = 1/cos(θ)
cot(θ) = 1/tan(θ)

Notice how we can multiply things to get rid of fractions, too.

1 = sin(θ)csc(θ) = cos(θ)sec(θ) = tan(θ)cot(θ)

In addition to these, all the trig functions follow specific rules when they land in specific quadrants.  Or rather, they have specific signs (+ or -) in specific quadrants.  LISTING TIME!

θ in Quadrant    sinθ   cosθ   tanθ   cotθ   secθ   cscθ
      I                  +        +        +       +        +        +
     II                  +        -         -       -         -         +
     III                 -         -        +       +        -         -
      IV                -         +       -        -         +        -

How do we know that sinθ is ALWAYS negative in QIII?  Well, look at it.  sinθ = y/r, right?  r > 0 for ALL quadrants, so it's ALWAYS positive.  And then y is negative in QIII, so it's a negative divided by a positive, which is a negative.  You can apply the same logic to all the other functions in the other quadrants.

So, if we have cosθ = 2/3, then we know it is either in QI or QIV because the value is positive. 

Now, let's talk about the range of these functions.  You can't always just say sinθ will = any number you want.  There are a few rules.  This actually simplifies things, if you think about it.

Look at the triangle we drew last section with x, y, and r segments.  As we rotate the segment r upwards until it is perfectly vertical, we see that the value of y changes, getting larger and larger, but never getting larger than r itself.  It can, however, = r.  So, we say y < r.

If we divide by r on both sides, we see y/r < 1.  If we do the same, only this time we rotate r clockwise so it goes through QIV first, we ultimately get to see that -y < r, which changes to -y/r < 1, which is the same as y/r > -1.

We put the two inequalities together.  -1 < y/r < 1.  What is y/r?  It's SINE!!  So, we can say -1 < sinθ < 1.

...Similarly, we can do this with cosθ, and we have different ideas with secθ and cscθ.  Ultimately, though, these are our ranges:

-1 < sinθ < 1               ,  -1 < cosθ < 1
secθ > 1 or secθ < -1  ,  cscθ > 1 or cscθ < -1
tanθ = anything            ,   cotθ = anything

Now, we only have two more things.  Quotient Identities and Pythagorean Identities.

Quotient Identities.  Let's just randomly decide to throw sinθ / cosθ.  sinθ = y/r, cosθ = x/r.
sinθ/cosθ = (y/r)/(x/r) = (y/r)*(r/x) = y/x = tanθ.  WHOAMG.

We have two of these.

sinθ / cosθ = tanθ, cosθ =/= 0
cosθ / sinθ = cotθ, sinθ =/= 0

Obviously, the denominator cannot be 0 or we get division by 0.  But those are it!  That just leaves...

Pythagorean Identities.  Remeber r² = x² + y².  Let's divide everything by r².

r²/r² = 1 = x²/r² + y²/r² = (x/r)² + (y/r)².  What is x/r?  cosθ!!  What is y/r?  sinθ!!  Put 'em in!!

(cosθ)² + (sinθ)² = 1  -->  cos²θ + sin²θ = 1.

If you do the same process, but instead divide by x² or y², you ultimately get these three Pythagorean Identities (you should go ahead and work out the division by x² and then y² for some practice!):

sin²θ + cos²θ = 1
tan² + 1 = sec²θ
1 + cot²θ = csc²θ

Now for examples!

1)  Find the tangent of cotθ = -3.

Well, tanθ = 1/cotθ, therefore tanθ = 1/-3 = -1/3.

2)  If cscθ = 3, what is sinθ?

Well, sinθ = 1/cscθ, so sinθ = 1/3.

3)  Give the sign of the sine, cosine, and tangent for the angle 406^o.

Well, 406 > 360, so we subtract 360.  406 - 360 = 46^o.  46 lies in QI.  Therefore, all the signs are positive.

4)  Find the value of θ for
tan(3θ - 4^o) = 1/cot(5θ - 8^o).

Well, remember, tanθ = 1/cotθ, which we can just say tanA = 1/cotA.  Since they must be equal, and the trig functions are reciprocals, we ignore EVERYTHING and just look at the stuff on the inside, then set it = each other.  That is,

3θ - 4 = 5θ - 8.  Simplify -->  4 = 2θ.  θ = 2.

5)  Find secθ if tanθ = Sqrt(7)/3, where θ is in QIII.

Okay, we see a secθ and a tanθ in the same thing.  We can't use quotient or reciprocal because the two aren't related at all.  However, they ARE related in the Pythagorean Identity, tan²θ + 1 = sec²θ.

Plug stuff in.  (tanθ)² + 1 = (secθ)².  Secθ = Sqrt(7)/3, (secθ)² = (Sqrt(7)/3)² = 7/9.

So, we have (tanθ)² + 1 = 7/9.  Subtract 1, or 9/9, from each side.  (tanθ)² = (7/9) - (9/9) = -2/9.

So, (tanθ)² = -2/9.  We can't take the Sqrt of this, so let's just use the positive value for now (it's okay, as we'll learn later.  Ultimately, if you get a negative and need to take a square root, just turn it positive).  Sqrt(2/9) = Sqrt(2)/3.

tanθ = Sqrt(2)/3.  Now, we simply use a calculator to find the angle of it by hitting 2nd TAN (on a TI calculator) to get tan^-1(tanθ) = tan¹(Sqrt(2)/3).  tan¹(tan) cancels out, so we have θ = tan^-1(Sqrt(2)/3) = 25.23940182.

However, we were told to find it in QIII.  To do this, we simply add 180 degrees to this.  25.239 + 180 = 205.239^o.


--------


What if we were asked to find QI?  Well, the 25 < 90, so it's already in QI.  What if QII?  We just do 180 - number.  What if QIV?  We do 0 - number.  Keep that in mind for other problems.


Why do we care?  ALL those rules help simplify crap super quickly when we need to, and they help us reach answers we otherwise couldn't reach because, really, we just wouldn't know.  However, the most important things are the Pythagorean Identities.  MEMORIZE THEM.  They come up all the time in later math classes in helping to reduce a LOT of crap (especially when you see a sin²θ + cos²θ under a square root sign, and if you know that that = 1, then you just take the square root of 1, which is 1.  VERY handy).

Now we should actually introduce the functions we'll be using from here on out.  They are, of course, trig functions.  They consist of the sine, cosine, trangent, secant, cosecant, and cotangent.

If we draw a point on our origin (0,0), then move to the right some x distance and move up some y distance, then draw a line segment r from the y's endpoint down to the origin, we have a triangle.  We can measure the length of r using the distance formula, r = Sqrt((x - x₀)² + (y - y₀)²), where x and y are arbitrary distances, x₀ and y₀ are the original points of (x,y), which in our case is the origin, and r is the length of the hypotenuse.

Because x₀ = y₀ = 0, we can simplify the above to Sqrt[(x-0)² + (y-0)²] = Sqrt(x² + y²).

Now, so we have the values of x, y, and r.  Now we can define the 6 trig functions.  First of all, looking at this triangle, the little angle between the x segment (the bottom, horizontal segment) and the diagonal slant (r) makes an angle we will just call θ.

Now, let's define.

Sine:  sin(θ) = y/r, r =/= 0.     =/=  means "is not equal to."
Cosine:  cos(θ) = x/r, r =/= 0.
Tangent:  tan(θ) = y/x, x =/= 0.
Secant:  sec(θ) = r/x, x =/= 0.
Cosecant:  csc(θ) = r/y, y =/= 0.
Cotangent:  cot(θ) = x/y, y =/= 0.

Obviously, we can't have 0 in the denominator or else that's division by 0, which is bad.

Also, note something about the tangent, tan(θ) = y/x.  Haven't you seen y/x somewhere before?  Aha!  Slope!  IN GENERAL, tan(θ) = m, so the tangent is the same as the slope.  Not ALWAYS, but USUALLY.

Now for a simple table before we do some examples.  Remember quadrantal angles?  Those things on the XY axes themselves?  0^o, 90^o, 180^o, 270^o, 360^o?

It's good to just memorize this table (you ultimately, as you work in Trig, just come to naturally memorize them because you will use them a lot.  As well as something called the Unit Circle, which we will tackle later).

Anyway, here's our table:

θ       sinθ    cosθ    tanθ           cotθ       secθ        cscθ
0        0        1         0          Undefined     1        Undefined
90      1        0      Undefined      0        Undefined    1
180    0        -1         0         Undefined    -1       Undefined
270   -1        0      Undefined     0        Undefined    -1
360    0         1         0         Undefined     1         Undefined

Also, a huge note:  UNTIL WE HIT CHAPTER 3, KEEP YOUR CALCULATOR IN DEGREE MODE.  For TI calculators, just hit the Mode button, then scroll until you see the row that says Radians or Degrees, and highlight Degrees and hit Enter, then 2nd Mode to Quit.

Now for some examples!!

1)  Find the 6 trigonometric values for the point (x,y) = (0,2).  Rationalize the denominator if applicable.

Okay, we have x = 0, y = 2.  Now we need r.  Remember the first thing?  r = Sqrt(x² + y²) = Sqrt(0² + 2²) = Sqrt(0 + 4) = Sqrt(4) = 2.  You may think you need +2, but you are wrong.  In this case, our "r" is a "distance," which is ALWAYS measured with positive values.  Later, we will see negative r's, but so long as you are taking square roots of things, you stick to positive values.

So, x = 0, y = 2, r = 2.  Now we just do the trig functions:

sin(θ) = y/r = 2/2 = 1.
cos(θ) = x/r = 0/2 = 0.
tan(θ) = y/x = 2/0 = undefined.
csc(θ) = r/y = 2/2 = 1.
sec(θ) = r/x = 2/0 = undefined.
cot(θ) = x/y = 0/2 = 0.

Bam, done.

2)  Given that the quadrant your point (x,y) is in quadrant IV, decide if x/y is positive or negative.

Well, quadrant IV is the bottom-right.  This means x = positive (right), y = negative (bottom).  So, we have +/-, which is negative.

3)  Use the table of quadrantal angles to evaluate the expression:  tan360^o + 4sin180^o + 5cos²180^o.

According to our table, tan360^o = 0, sin180^o = 0, and cos180^o = -1.  For sin180, we multiply this by 4 because we have 4sin180, so 4*0 = 0.  For the cos², that is the same thing as writing cos(θ)cos(θ).  So, we know cos180 = -1, and we just multiply that by cos180, which is also -1.  -1 * -1 = +1.  Then we multiply that by 5.  5 * 1 = 5.

Now put them all together.  0 + 0 + 5 = 5.

4)  If n is an integer (a non-decimal, non-fraction number including negatives, 0, and positives), n*180^o represents an integer multiple of 180 degrees, and (2n + 1)*90 degrees represents and ODD integer expression.  Decide if the expression is = 0, -1, 1, or undefined;  cos[(2n + 1)*90].  (note:  Mindsay has failed and refuses to let me make the degree symbol anymore for this entry.  As a result, assume the 90 is 90 degrees, while the rest are just 2n + 1).

Okay, we have cos[(2n + 1) * 90 degrees].  Look at our table for what cos(90) is.  It is ALWAYS 0.  Now, no matter WHAT integer we put in for n, we will ALWAYS get an odd number out.  2*any integer is always an even number, right?  Add 1 to it, and you ALWAYS get an odd.  So, we ALWAYS get an odd * 90 degrees.

What is an odd*90 degrees.  Let's try some odd numbers, like 1.  1*90 = 90.  3.  3*90 = 270.  5*90 = 450, and so on.  Note that 450 is > 360, so we should do 450 - 360 = 90 to get an idea of where this value is.

If we do it infinitely, we constantly alternate between 90 and 270 degrees.  The cosine at EITHER of these angles is ALWAYS 0.  Thus, our answer is 0.

---------------

Why do we care?  Dude, I'm not really even going to try.  You need to know the 6 trig functions because, hello, this is TRIGONOMETRY.  We will be using them EXTENSIVELY, and about NONSTOP pretty soon.  Best to learn to love them.

Draw 2 lines, m and n, that do not intersect and move horizontal, so that m is higher up than n.  Two or more lines that lie in the same plane and never intersect are called parallel lines.  Now, draw another line that goes right through them to the up-right/down-left, label this line q.  Line q is called a transveral line, that is, a line that intersects two parallel lines.

Notice how at each intersection, 4 angles are created, totaling 8 angles.  From the top to bottom, left to right, label the angles 1, 2, 3, 4, 5, 6, 7, 8.  Each of these angles has a lot of interesting properties with another.  Hopefully you have a big drawing, 'cuz there are a LOT of things to say here.

Angles 1, 4, 5, and 8 are all equal.  Angles 2, 3, 6, and 7 are all equal. 

We call angles 4 and 5 alternate interior angles.  Likewise, 3 and 6 are also alternate interior angles.

We call angles 1 and 8 alternate exterior angles.  Likewise, 2 and 7 are also alternate exterior angles.

We call angles 2 and 6 corresponding angles.  Likewise, angles 4 and 8, 7 and 3, and 5 and 1 are all corresponding angles with each other.

We call angles 2 and 3 vertical angles.  Likewise, 1 and 4, 5 and 8, and 6 and 7 are all vertical angles.

Why do we call them these angles?  Well, let's look at the vertical angles.  Vertical angles are ALWAYS equal, and then they lie, basically, as mirrored reflections of one another.  Don't those angles look mirrored?  Good.  Now, let's look at corresponding angles.  We call them that because, basically, they are equal and basically in the same position as each other, just "shifted" or translated up, right, down, or left.  They still look exactly the same, so they correspond with each other.

Now let's look at alternate exteriors.  With 1 and 8, we see 1 is the top-left of one intersection, and 8 is the bottom-right of the other intersection.  They are on different sides (left and right; top and bottom), and they are around different intersections, so they alternate, but are strictly on the outside of the transversal.

Alternate interior angles work the same way.  They are on different sides (left and right; top and bottom) and on different intersections.  However, they lie on the INSIDE of the transversal.

And for the love of God, keep that picture handy.  Redraw it if you have to, because I'm going to be, in my examples, saying "angle 1 is blah" and whatnot, and you need to have that on you so you know what I'm talking about.

Like now!  Let's say angle 1 = (3x + 12)^o, and angle 8 = (5x + 10)^o.  What is the value of x?

Well, we know angles 1 and 8 are alternate exteriors, so they are equal.  Thus, we can set their degrees equal to each other.  That is, 3x + 12 = 5x + 10.  Now we simply find x.  12 = 2x + 10.  2x = -2.  x = -1.

Now, for triangles.  We all know what triangles are, right?  They are three line segments, connected at the verteces.  Here is a very important rule about triangles:

The sum of the measures of the angles of any triangle is 180^o.  So, if you are told that one angle is 40^o, another is 21^o, then to find the third, you simply do 180 - 40 - 21 = 180 - 61 = 119^o.

Now, there are six kinds of triangles.

Acute Triangles have all angles < 90^o.
Right Triangles have exactly one angle = 90^o.  Note how it is impossible to have more than one angle = 90^o, because just having two would already give us 90 + 90 = 180^o, and so we couldn't fit a third angle.
Obtuse Triangles have exactly one angle > 90^o.  Note that it is impossible to have more than one angle > 90^o, or else we'd already have, after 2 angles, something > 180^o.
Equilateral Triangles have all angles = each other and all sides = each other.  In fact, ALL angles of an equilateral triangle must = 60^o, because triangles have 3 angles, and if they are equal, we have 180/3 = 60.
Isosceles Triangles have two sides equal.  Also, the angles opposite these two sides are equal to each other.
Scalene Triangles have no equal sides.

Now, we have two more things to discuss before working some problems.  The first is the idea of congruent triangles.  If you took a triangle, copied it precisely, and simply rearranged it so maybe the long end was in another direction, this is a congruent triangle.  Congruent triangles are all equal to each other, but they're placed differently.  One may be pointing up, another pointing down, another backwards, another upside-down, whatever.  As long as all their properties are equal.

The last thing is the idea of similar triangles.  Similar triangles have exactly the same shape, but not necessarily the same size.  I didn't say NEVER the same size, so all congruent triangles ARE similar triangles. 

Draw triangle ABC such that we have one segment going horizontal, and the other two go up-right and down-right.  Label, from left to right, the angles A, B, C.

Draw another triangle following the same procedure, but make it larger, and label its angles D, E, and F the same way I said.

We say angle A corresponds to angle D, B corresponds to E, and C corresponds to F.  Segment AB corresponds to DE, BC corresponds to EF, and AC corresponds to DF.

In order for these two triangles to be similar, 2 important rules MUST BOTH be fulfilled.

1)  Corresponding angles must have the same measure.
2)  Corresponding sides must be proportional (AB/DE = BC/EF = AC/DF).

Now let's do some examples!!

1)  If one angle of a triangle is 37^o and another is 52^o, what is the measure of the third angle?  What kind of triangle is this?

Well, triangles have measures of 180, so to find the third, we do 180 - the other two, or 180 - 52 - 37 = 180 - 89 = 91^o.  Note how one angle > 90, so therefore this is an obtuse triangle.

2)  Suppose the angle of a triangle is (x + 20)^o, another is x^o, and the third is (210 - 3x)^o.  What is the value of x?

Again, the triangle's angles sum must = 180.  So, we just sum all 3 angles and set it = 180, even if it looks all complicated and stuff.  Let's do it.

(x + 20) + (x) + (210 - 3x) = 180.  Simplify--   220 - x = 180.  220 = 180 + x.  x = 40.

3)  A tree casts a shadow 45 m long.  At the same time, the shadow cast by a vertical 2-m stick is 3 m long.  Find the height of the tree.

This is a similar triangle problem.  We see the base of one segment is 45 meters long, and the base of another is 3 meters long.  These two bases correspond to each other.  Now we have the height = x, and then the height of the stick = 2.  These two bases correspond.  We ASSUME the angles correspond (it is possible that the sun could be at a weird angle and thus ruin all this, which this book failed to mention).

So, we take the corresponding sides and make the ratio equal.  That is,
height of tree = height of stick
base of tree       base of stick

Or   x / 45 = 2 / 3.   Now we just solve for x.  3x = 45*2 = 90.  x = 30, thus the tree is 30 meters high.

4)  Draw a really big triangle, and make its bottom horizontal.  Now, somewhere near the top of the triangle, but still inside it, draw another horizontal line.  ALWAYS triangles like these are similar.  So, we know these triangles are similar.  Now, then.

The top-left tiny section has a length of x - 2y.  The top-right tiny section has a length of 5.  The little line you drew through the middle has a length of x - 5.  The bottom-right length has a length of 10.  The bottom length has a length of 15.  The bottom-left length has a length of x + y.  Your job is to find what x and y are.

Okay, now that we've hopefully drawn this correctly, let's remember our rules for similar triangles.  We gotta set the ratios equal to each other.  The first thing we're going to do is pick the two right segments.  Why?  'Cuz those have DEFINITE lengths.  They are 5 and 10.  No need to find anything.  Now, which next pair should we pick?  Well, if we pick any of the left ones, we are stuck with x and y, something we can't solve in one equation.  So, we pick the bottom ones, x - 5 and 15.  Now we simply set up our ratio.

x - 5  =  15
 5           15    Why did I use 15 for the denominator on the right?  It should be 10, shouldn't it?  NO.  The bottom segment is associated with the WHOLE BIG triangle, so the right segment must be associated with the WHOLE BIG triangle.  It is broken into 5 and 10, so we simply add them up to find the length of the WHOLE section.  Now we just solve for x.

(x - 5)/5 = 1.  x - 5 = 5.  x = 10.

Now, then.  We found x, so let's find y.  We're gonna use the 5 and 10 sides again because they are definite, and NOW we can pick the left sides.  Let's set up our ratios:

x - 2y   =  (x + y) + (x - 2y)
   5                 15

Now we simplify the numerator on the right.  x + y + x - 2y = 2x - y.

Then we do cross-multiplication to get rid of those icky fractions.  Just erase the denominators and multiply the left side by 15 and the right side by 5:

15(x - 2y) = 5(2x - y).  Now simplify.  15x - 30y = 10x - 5y.  -->  5x - 30y = -5y.  -->  5x = 25y.

Now, we already know x = 10, so plug it in!!  5(10) = 25y.  50 = 25y.  y = 2.

NOTENOTENOTE!!  You can flip the ratios however you want, as long as you stay consistent.  If you use the left side for the top on the left part of the = sign, you MUST use the left side of the whole on the right side of the = sign.  You can't use the left in the numerator and then the left in the denominator.  That's not a RATIO.  That's just randomly throwing crap together.

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Why do we care?  Similar triangles are very important to us.  They let us see great things we can't normally see.  ....Like the distance of the Sun.  We can't see how far away it is, but by using shadows and measuring angles, we could, in the sand, draw a similar triangle.  Then we simply set up a series of ratios to find how far away from the Sun we are.  ...It was a little more complicated than that, but once you remove the other variables, this was basically what was done.

Triangles are important in all sorts of things.  They also allow us to understand the Greek letter pi, 3.1415 and so on.  They also let us figure things out in circles, which we'll be doing later on.

We gotta define some words here.  Don't worry.  They're simple.

Draw two points.  Label one A and the other B.  Draw a line through them, and draw arrow heads at the end of each line.  This is called line AB.  The portion consisting ONLY of A and B and everything in between is called a line segment, in this case, segment AB.  Take only point A, draw a line STARTING from A and hits B then goes THROUGH B, with an arrow head on the end.  This is called ray AB.

A line has infinite length and NO endpoints.  A line segment has finite length, the distance from A to B and has TWO endpoints.  A ray has ONE endpoint and infinite length.

An angle is formed by rotating a ray about its one endpoint, in our case, A.  You can also think of it as two distinct rays that go off in different direction, and the angle is the measurement BETWEEN those two rays.  If you rotate the ray counterclockwise, the angle measurement is positive.  If you rotate it clockwise, its measurement is negative.  If you rotated a ray 30 degrees counterclockwise, you have an angle of 30 degrees.  If you rotated it 30 degrees clockwise, you have an angle of -30 degrees.  Got that?  We use denote these as n^o for degrees, where n is any number.  18, 90, whatever.  The important thing is that by tacking the little o thing, we are being told it is a degree.

Now, the ray's initial position, almost always horizontally and pointing to the right because that's the convenience most mathematicians and textbooks use, is called the initial side.  When you rotate it, the new position is called the terminal side.  The endpoint of the ray is called a vertex.

A measurement of an angle, for right now, is called a degree (later, we will discover radians).  You probably know this already, though.  If the angle < 90 degrees, it is called an acute angle.  If = 90 degrees, it is called a right angle.  If > 90 degrees, it is called an obtuse angle.  If exactly 180 degrees, it is called a straight angle.

If you sum two angles and you get exactly 90 degrees, the two angles are complementary angles.  Say, 45 degrees + 45 degrees.  Or 30 + 60.  Or 80 + 10.  Whatever.  If the two angles sum to be 180 degrees, they are supplementary angles.  179 + 1.  Two right angles (90 + 90).  Whatever.

Now, a degree isn't the only way we measure angles.  Sometimes we need decimals in our angles.  What if you have 20.787878 degrees?  Well, can you actually visualize that?  Probably not.  Actually, you probably won't be able to visualize these next two terms, but you need them for calculation and conversion purposes.

The first "decimal measure" is called the minute.  We denote this as n', where n is just some number. ANY number.  The important thing is that we use ONE apostrophe to tell us it is a minute.  Do NOT confuse this with the TIME "minute," which is 60 seconds.  The GEOMETRIC minute is simply 1/60 of a degree.  If you have 2 degrees, you have 2 / (1/60) = 120 minutes.    It takes 60 minutes to make 1 degree, so, to convert, you simply do

x degrees   *   60 minutes
      1                1 degree

The word "degrees" cancels out (we are actually cancelling units), and the only word that remains is "minutes."  Likewise, to convert from minutes to degrees--

x minutes  *  1 degree
     1             60 minutes

The next one is called a second.  DO NOT CONFUSE WITH THE TIME SECOND.  A second is 1/60 of a minute.  We denote this by n'', where n is some number, ANY number.  We use TWO apostrophes here.  Our conversions are--

x degrees   *   60 minutes   *   60 seconds
    1                   1 degree           1 minute

x seconds  *   1 minute     *   1 degree
    1                60 seconds    60 seconds

If you only need to go from seconds to minutes, just drop the 1 degree / 60 seconds on the end of that second one there.

Remember when I said that most mathematicians and textbooks have the rays pointing to the right?  There's a term for this.

If, on a graph, the vertex is at the origin (0,0) and the ray points perfectly straight to the right along the x-axis, we call this the standard position.

We then rotate this angle throughout the four quadrants of the graph.  Quadrant I is the top-right part of the graph.  Quadrant II is the top-left.  Quadrant III is the bottom-left.  Quadrant IV is the bottom-right.  KNOW THIS.  See your actual axes?  The XY axes?  Those "lines" actually have specific degree measurements.  The positive-y (pointing up) is 90 degrees.  Negative-x (pointing left) is 180 degrees.  Negative-y (pointing down) is 270 degrees.  And positive-x (pointing right) is 0 or 360 degrees.  Each of these four (actually five if you count both 0 AND 360) "angles" is called quadrantal angles.  This is because they DIVIDE the four quadrants.

If you make a COMPLETE rotation, that is, go a full 360 degrees so you're back in the standard position, you are a coterminal angle.  ALL angles > 360 have at least ONE coterminal angle.  To find just where the ray points if it's > 360, simply subtract 360 from the angle.  For example...

Suppose we have an angle of 450 degrees.  What is this?  Well, let's subtract 360 from 450.  450 - 360 = 90.  Bam.  Our ray now falls on the 90 degree angle!!  See how easy that was?

Let's do some examples!!

1)  62^o 18' + 21^o 41'.

All we're doing is adding things SEPARATELY.  Remember, a minute is NOT a degree.  It's a PART of a degree, so you can't just carry numbers over together.  Let's add the minutes first.  41 + 18 = 59.  That's not 60, so we're okay.  Now let's add the degrees.  21 + 62 = 83.  So, our final answer is 83^o 59'.

2)  180^o - 124^o 51'.

We don't have any minutes with the 180.  However, we remember that one degree is 60 minutes.  So, why don't we take off one degree so we can have 60 minutes?  180 degrees - 1 degree = 179 degrees.  Now we have these 60 minutes in there, too.  Put them together!  179^o 60'.  Remember, if you don't have any units you need, just take one away from the units you HAVE to get 60 of the units you need (assuming what you need is smaller than what you have.  Minutes are smaller than degrees).

Now instead we have 179^o 60' - 124^o 51'.  Now we can subtract!  Subtract separately, now.  179 - 124 = 55.  60 - 51 = 9.  Our final answer is 55^o 9'.

3)  Find the angle of smallest measure coterminal with the angle -40^o.

What does this mean?  We know the coterminal angles are 0, 90, 180, 270, and 360.  -40 is in Quadrant IV, so we know it has to be either 0 or -90 (-90 is the same as 270.  We measured counterclockwise for being positive, so we can measure clockwise for negative).  Well, -40 is closer to 0 than it is to -90, isn't it?  So, our answer is 0^o.

4)  Convert 31.4296^o to degrees, minutes, and seconds.

Well, we know we can just take the 31^o by itself.  Now we're left with .4296^o we need to convert.  Why?  Because minutes and seconds are FRACTIONS of a degree, that is, they are < 1 degree.  This decimal < 1, right?  So, yeah, we only worry about it for degrees.

If we remember, to get from degrees to minutes, we multiply by 60 (scroll up).  So.. 0.4296 * 60 = 25.776.  Now we can hack off the 25 because it's some minutes, leaving the decimal to be converted into seconds (because a second is a FRACTION of a minute, just like what we did a second ago).  Now, to convert from minutes to seconds, we multiply by 60 again.  0.776 * 60 = 46.56.

Let's put everything together.  31^o 25' 46.56".

5)  A pulley rotates through 75^o in 1 minute.  How many rotations does it make in 1 hour?

A word problem!  We have 75^o per 1 minute, and we have 1 hour.  Let's do some CONVERSIONS!

75 degrees     *    60 minutes    *   1 revolution   =  12.5 revolutions/hour.
    1 minute                1 hour           360 degrees

Whoah, how did I do all that?  You start with what you're given.  We were told we have 75 degrees in one minute.  That also means per minute.  So, we have our first fraction.  We know there are 60 minutes in 1 hour, so we set up the next fraction as a way to "cancel out" the word "minutes."  It WAS in the denominator, so we put it in the numerator because anything / itself = 1.  We did the same thing with the degrees in the NEXT one (though, you could have done this one first).  360 degrees are in 1 revolution.  And then it's just multiplying.

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Why do we need to know about angles?  Well, geez, why not?  If you don't know anything about angles, you can't do the real trig stuff later LOLOLOL.  Also, the key point here was how angles work.  They are measured ROTATIONS about the vertex.  CD players rely on this.  CD players rotate CDs at a very specific speed.  You need to know this speed in order to determine how much information CDs can store and at what layering so the CD player's speed can read it correctly and not skip or scratch the disc.