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 Part II A couple of more examples of using Trig functions to solve the sides of a triangle.
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- Welcome back.
- I just want to do a couple more of these problems, just so you,
- I guess, see a few couple more problems and get some practice.
- So let's start with another problem.
- As always, let me draw my triangle.
- I always like to draw it a little different
- to confuse you.
- Let's say that this angle up here is 0.36 radians.
- An angle is 0.36 radians, and then this side right here, this
- top side-- let me do it in a different color-- this top side
- in pink is square root of 73 units long or whatever.
- It could be inches or feet or whatever.
- But it's square root of 73 units long.
- So my question is, what is this side here, this green side?
- Well, how did we do this type of problem last time?
- We should figure out what sides we're dealing with.
- Are we dealing with the adjacent and the opposite, the
- adjacent and the hypotenuse, the opposite and the
- hypotenuse, and then we'll know which trig functions we
- should be dealing with.
- So we know this side, this pink side-- and this pink side, it
- should be almost second nature by now, is the
- hypotenuse, right?
- It's the longest side, and it's opposite the right angle.
- So we know the hypotenuse, and what are we solving for?
- Well, this is the angle we know, so we're solving
- for the opposite side.
- So we're solving for the opposite side, and we know the
- hypotenuse, so what trig function will we probably use?
- Let's write out our mnemonic.
- SOH CAH TOA.
- So what did I say before?
- We're solving for the opposite sign, and we know the
- hypotenuse, so we are going to use the opposite and
- the hypotenuse.
- So which of these is that?
- The opposite and the hypotenuse, the o
- and the h, right?
- It's SOH, right?
- And SOH says that sine is equal to the opposite
- over the hypotenuse.
- Sine of an angle-- in this case, it's this angle-- sine of
- 0.36 radians-- remember, this is radians we're dealing with,
- not degrees-- is equal to the opposite side.
- And the opposite side is this green side right here, so I'll
- just write opposite instead of writing o because o would look
- like a 0-- is equal to the opposite side over the
- hypotenuse, right?
- This is just sine is equal to opposite over hypotenuse.
- Well, what's the hypotenuse length?
- Well, it's the square root of 73.
- These are p's, by the way.
- I know they don't look like p's, but opposite.
- So the opposite side-- we're just multiplying both sides by
- the square root of 73-- is equal to the square root of 73
- times the sine of 0.36 radians.
- Now, once again, I don't know what the sign of 0.36 radians
- is in my head, but I'll tell you the answer.
- The sine of 0.36 radians is equal to-- I'm just rewriting
- this-- the sine of 0.36 radians, if you looked it up on
- a table or if you used your calculator in radian mode, is 3
- square roots of 73 over 73.
- And, of course, your calculator is going to give you
- something-- it'll give you some decimal number.
- I won't write it this way.
- So just remember, this is this.
- And I just looked that up.
- There's no magic there.
- And the square root-- or you could use a calculator.
- On the Khan Academy, when you do problems, it'll actually
- tell you what it is, so you don't have to use a calculator.
- So now we just simplify.
- Square root of 73 times square root of 73 is 73 over 73 is
- equal to 1, so these all cancel out.
- And we get the answer of 3.
- So this side right here is 3.
- And just out of curiosity, if you wanted to solve for this
- side, there's two ways we could do it, right?
- We could use the Pythagorean theorem, because, you know, a
- squared plus b squared is equal to c squared, or we
- could use trigonometry.
- I'll let you guess what trig function--
- actually, let's do that.
- Let's figure out that side using trigonometry, and then
- let's figure out that side using the Pythagorean theorem,
- just to show that everything fits together in math.
- So I wrote that 3 there, so I can erase all of this stuff.
- Let me erase it.
- I shouldn't have erased the SOHCAHTOA.
- Actually, we should have that memorized by now.
- SOHCAHTOA.
- All right, so let's figure out what this orange side is here.
- And if you think about it, we could do it a bunch of ways.
- We could say, well, this is the adjacent side, right?
- Because we know the opposite and we know the hypotenuse.
- So we could either use-- we know the opposite, so we could
- say what trig function uses the opposite and the adjacent?
- Well, that's tangent function, right?
- So we could say tangent of 0.36-- let's call
- this side A, right?
- A for adjacent.
- Tangent of 0.36 is equal to the opposite, 3, over
- the adjacent, over A.
- Is there another trigonometry way we could think about this?
- Well, we also know the hypotenuse.
- What trig function uses the hypotenuse and the adjacent?
- Well, if you remember, SOH CAH TOA.
- CAH, cosine is adjacent over hypotenuse.
- So we could say cosine of 0.36 is equal to adjacent
- over square root of 73.
- And I'll just write SOHCAHTOA here, just so you can
- confirm what I'm doing.
- The TOA says the tangent is equal to the opposite, 3, over
- the adjacent, and CAH tells us the cosine is equal to the
- adjacent over the hypotenuse.
- So we could solve either one of these.
- If we use the second formula, we would get the adjacent side
- is equal to the square root of 73 times the cosine of 0.36,
- and then if you use your calculator in radian mode, or
- I'll just tell you, that cosine of 0.36 radians is equal to 8
- square roots of 73 over 73.
- And you can confirm that by getting a decimal number and
- then-- making sure once again your calculator is not in
- degree mode, but in radian mode.
- I think that's actually the default mode in a lot of
- calculators-- and solving for this.
- But once again, the 73, this square root of 73 times this
- square root of 73, is equal to 73, and then divided by 73.
- These all cancel out.
- How convenient, huh?
- And you get 8.
- So the adjacent side is equal to 8.
- And so if we'd solved for A here using the tangent
- function, we should've also gotten the adjacent
- side is equal to 8.
- And just to show you that everything works out from other
- concepts, let me show you this using the Pythagorean theorem.
- So 8 squared plus 3 squared should equal
- the hypotenuse squared.
- The square root of 73 squared.
- Well, 8 squared is 64 plus 9 should equal-- what's the
- square root of 73 squared?
- Right, it's 73.
- And, of course, 64 plus 9 is 73.
- And sure enough, that equals 73, so it works.
- Isn't that interesting how math just kind of fits together?
- I think at this point you're ready to try the modules, the
- Trigonometry II modules, and I guess let me know if you
- have any problems, or if you want to see more videos.
- 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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