Basic trigonometric ratios
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Basic Trigonometry
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Trigonometry 0.5
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Basic Trigonometry II
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Trigonometry 1
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Example: Using soh cah toa
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Example: The six trig ratios
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Reciprocal trig functions
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Example: Using trig to solve for missing information
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Trigonometry 1.5
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Example: Calculator to evaluate a trig function
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Example: Trig to solve the sides and angles of a right triangle
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Trigonometry 2
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Example: Solving a 30-60-90 triangle
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Special right triangles
Using Trig Functions Using Trigonometric functions to solve the sides of a right triangle
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- We're now going to do a few examples to actually show
- you why the trig functions are actually useful.
- So let's get started with a problem.
- Let's say I have this right triangle.
- That's my right triangle.
- There's the right angle.
- And let's say I know that the measure of this angle
- is pi over 4 radians.
- And I'll just write rad for short.
- If the measure of this angle is pi over 4 radians, and I also
- know that this side of the triangle-- this side right
- here-- is 10 square roots of 2.
- So I know this side of the triangle.
- I know this angle, which is pi over 4 radians.
- And now, the question is, what is this side of the triangle?
- I'm going to highlight that.
- And let me make it in orange.
- So let's figure out what we know and what we
- need to figure out.
- We know the angle, pi over 4 radians.
- And actually, turns out if you were to convert that
- to degrees, it would be 45 degrees.
- And we know-- what side is this?
- This is the hypotenuse of the triangle, right?
- And what are we trying to figure out?
- Are we trying to figure out the hypotenuse, the adjacent
- side to the angle, or the opposite side to the angle?
- Well, this is the hypotenuse, we already know that.
- This is the opposite side.
- This is the opposite side.
- And this yellow side is the adjacent side, right?
- It's just adjacent to this angle.
- So we know the angle, we know the hypotenuse, and we want to
- figure out the adjacent side.
- So let me ask you a question.
- What trig function deals with the adjacent side
- and the hypotenuse?
- Because we have the adjacent side is what we want
- to figure out, and we know the hypotenuse.
- Well, let's write down our mnemonic, just in
- case you forgot it.
- SOHCAHTOA.
- So which one uses adjacent and hypotenuse?
- Right?
- It's CAH.
- And CAH, the c is for what?
- The c is for cosine.
- Cosine of an angle-- let's just call it any angle-- is equal to
- the adjacent over the hypotenuse.
- So let's use this information to try to solve for this orange
- side, or this yellow side.
- So we know that cosine of pi over 4 radians-- so let's say
- cosine of pi over 4-- must equal this adjacent
- side right here.
- Let's just call that a. a for adjacent.
- The adjacent side divided by the hypotenuse.
- The hypotenuse is this side.
- And in the problem, we were given that it's
- 10 square roots of 2.
- So we can solve for a by multiplying both sides of
- this equation by 10 square roots of 2.
- And we will get-- because, right?
- If we just multiply times 10 square root of 2,
- these cancel out.
- And then you get a 10 square root of 2 here.
- So you get a is equal to 10 square roots of 2 times
- the cosine of pi over 4.
- Now you're probably saying, Sal, this does not look too
- simple, and I don't know how big the cosine of pi over 4 is.
- What do I do?
- Well, no one has the trig functions, or the values of
- the trig functions memorized.
- There's a couple of ways to do it.
- Either I could give you what the cosine of pi over 4 is.
- That's sometimes given in a problem.
- Or you can make sure that your calculator is set to radians
- and you can just type in pi divided by 4-- which is roughly
- 0.79-- and then press the cosine button.
- You finally know what it's good for.
- And you'll get a value.
- Or-- and this is kind of the old school way of doing it--
- there are trig tables where you could look up what cosine
- of pi over 4 is in a table.
- Since I don't have any of that at my disposal right now,
- I'll just tell you what the cosine of pi over 4 is.
- The cosine of pi over 4 is square root of 2 over 2.
- So a, which is the adjacent side-- a for adjacent-- is
- equal to 10 square roots of 2 times square root of 2 over 2.
- Remember, to get the square root of 2 over 2, you might
- be a little confused.
- You're like, how did Sal get that?
- All I said is, the cosine of pi over 4 is square
- root of 2 over 2.
- And that's not something that-- well, actually, this one you
- might know offhand, because of the 45 degree angle.
- But this isn't something that people memorize.
- This is something you would look up, or it's given in
- the problem, or you'd use a calculator for.
- And a calculator, of course, wouldn't give you square
- root of 2 over 2.
- It'd give you a decimal number that's not obviously
- square root of 2 over 2.
- But anyway, I told you that the cosine of pi over 4 is the
- square root of 2 over 2.
- And so if we multiply, what's the square root of 2 over 2?
- What's the square root of 2 times the square root of 2?
- It's 2.
- So that's 2, and then that cancels with that 2.
- And so everything cancels except for the 10.
- So the adjacent side is equal to 10.
- Let's do another one.
- Let me delete this.
- Give me 1 second.
- I'm actually-- this is one of the few modules that I'm not
- generating the problems on the fly, because I need to make
- sure that I actually have the trig function values
- before I do the problem.
- So let's say I have another right triangle.
- I probably shouldn't have deleted that last one.
- So let's see, this is my right triangle.
- How much time do I have-- about 4 minutes left.
- Should be enough.
- So this is my right triangle.
- And I know the angle-- let's call this--.
- I know this angle right here is 0.54 radians.
- And I also know that this side right here is 3 units long.
- And I want to figure out this side.
- So what do I know?
- Well, this side is what side relative to the angle?
- It's the opposite side, right?
- Because the angle is here, and we go opposite the angle.
- So this is the opposite side.
- And what's this side?
- Is this the adjacent side, or is it the hypotenuse?
- Well, this is the hypotenuse, right?
- The long side, and it's opposite the right angle.
- So this is the adjacent side.
- So what trig function uses opposite and adjacent?
- Let's write down SOHCAHTOA again.
- SOHCAHTOA.
- TOA uses opposite and adjacent.
- OA.
- So T for tangent, right?
- TOA.
- So tangent is equal to opposite over adjacent.
- So let's use that.
- So let's take the tangent of 0.54 radians.
- So the tangent of 0.54 will equal the side opposite to it.
- So that's 3, right?
- The opposite side is 3.
- Over the adjacent side.
- Well, once again, the adjacent side is what we don't know.
- So we have to solve for a.
- So if we multiply both sides by a, we get a tan of 0.54--
- we could do that because we know it's not 0-- equals 3.
- Or a is equal to 3 divided by the tangent of 0.54.
- So once again, I don't have memorized what the tangent of
- 0.54 is, but I will tell you what it is because you also
- don't have it memorized.
- Or you could use a calculator to figure it out if you
- had a radian function.
- The tangent of 0.54 is equal to-- let me make sure
- I have this right.
- Oh, right.
- The tangent of 0.54 is 3/5.
- So then a is equal to 3 over 3/5.
- Right, the adjacent side-- now, once again, how
- did I get this 3/5?
- Well, I just told you.
- Or you can use a calculator to know that the
- tangent of 0.54 is 3/5.
- And of course, I'm using numbers that work out
- well, just so that the fractions all cancel.
- So we know that the adjacent side is equal to-- when you
- divide by fractions, it's like multiplying by the numerator.
- Multiplying by the inverse.
- So times 5/3.
- So the adjacent side is equal to 5.
- There.
- There you go.
- So let's just think about what I always do.
- I think about what I have, what sides I have, and what
- side I want to solve for.
- And in this case, it was the opposite side I had, and I
- wanted to solve for the adjacent side.
- And I said, what trig function involves those 2 sides?
- The opposite and the adjacent.
- I wrote down SOHCAHTOA.
- I said, oh, TOA.
- Opposite and adjacent.
- That's tan.
- So I took the tan of the angle.
- And then I said, the tan of the angle is equal to the opposite
- side divided by the adjacent side.
- That's right here.
- And then I just solved for the adjacent side.
- And of course, I used a calculator, or I told you
- what the tangent of 0.54 is.
- I think I'll do a couple more of these problems in the next
- module, but I'm out of time for now.
- Have fun.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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