Trigonometry word problems (part 1) The first part of a problem when the captain of a ship goes off track.
Trigonometry word problems (part 1)
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- Now that we know some trigonometry, let's use
- that trigonometry to solve some word problems.
- So let's get started with-- This is exciting.
- This is a navigation problem.
- So I have a ship that needs to go from point A to point B.
- So let me draw that.
- So I'm going from point A, which is that
- point, to point B.
- Let me draw a nice, big-- well not so big --triangle.
- Big enough.
- There you go.
- Because I need space to do my math.
- So let me-- Let's see.
- So this is point A.
- Point A I will draw in mauve.
- Point A to point B.
- So this ship needs to go from point A to point B.
- And this problem tells us that the distance between point A
- and point B is 10 kilometers.
- So this guy set sail and, because of whatever reason.
- He's a bad navigator or the water is flowing
- faster maybe upwards.
- Maybe the water's flowing in this direction.
- Let me draw that in blue to show which direction the
- water-- Say the current is flowing in that direction.
- And he ends up off track.
- He actually ends up, after 5 kilometers-- So he
- travels for 5 kilometers.
- Let me do that in blue-green.
- He travels for 5 kilometers and after 5 kilometers, he realizes
- that he has gone off track.
- And somehow-- I don't know maybe using his astrolabe.
- I don't really know what one is, but that's what people use
- when they're on ships to figure out where they are I think.
- --he realizes that he is 15 degrees off course.
- So what that means is that he's-- Between the path he took
- and the path he should've took is 15 degrees.
- At least that's my interpretation of it.
- So this is 15 degrees.
- And, of course, he's traveled 5 kilometers in the
- wrong direction.
- 15 degrees in the wrong direction.
- So what he needs to know is-- So this is-- Let
- me draw a little ship.
- So this is where the ship is now.
- That's my ship.
- So now this guy, this poor poor captain, he said so how far do
- I still have to go to get to point B?
- So let me-- So he needs to know how far is-- let me pick a nice
- color --how far is this distance right here?
- This may just involve some trigonometry.
- So how can we figure out this distance?
- Well let's just break down what we know.
- And that's how I do trig problems.
- I just keep messing around with it until I get
- the right answer.
- So let's see if we can tackle this problem that way.
- So what can we figure out?
- We know this is 5 kilometers.
- This side is 5 kilometers.
- We know that this is 10 kilometers.
- We know this angle.
- Well one thing I know I can figure out is-- Let's just make
- some right angles so we can start using some trigonometry.
- Let me draw a right angle here.
- I just drop it straight down.
- So can I figure out what this side is right here?
- Let me draw that in a different color.
- In this brown color.
- Can I figure out what this side right here is?
- Well this side is going to be what?
- This is-- If we look at this angle, this is adjacent to
- the angle and this is the hypotenuse.
- So if we know the hypotenuse, we know the angle, we want to
- figure out adjacent, what trig function should we use?
- Let me write out-- down our SOH CAH TOA.
- So we know the hypotenuse.
- We want to figure out the adjacent.
- So what involves adjacent and hypotenuse?
- CAH, or cosine.
- So we know that the cosine of 15 degrees-- so I'm going to
- write over this just not to waste space --cosine of 15
- degrees is equal to this brown side.
- So let's just call this-- I don't know.
- I'm just going to call it x, right?
- So this is equal to x over the hypotenuse, over 5, right?
- If we solve for x, we get x is equal to 5
- cosine of 15 degrees.
- Maybe we made progress.
- Maybe we didn't.
- Let's keep trying.
- All right.
- Now let's see if we could figure out this side
- of this triangle.
- Well this is, to this angle, this is the opposite, right?
- This is the opposite.
- And we know the hypotenuse.
- So what trig identity should we-- trig
- function should we use?
- well if we know the opposite-- Well if we want to figure out
- the opposite and we know the hypotenuse and if
- we know the angle.
- So what trig function deals opposite-- I was going
- to say hypotenuse.
- But anyway.
- What trig function deals with the opposite of the hypotenuse?
- Well that's sine, right?
- So the sine-- We call this y.
- We know that y is equal to-- Well, doing it the same way,
- sine-- 5 sine of 15 degrees.
- I skipped a step, right?
- Because we could say that sine of 15 degrees is equal to the
- opposite, y, over the hypotenuse, over 5.
- And then that step takes us here.
- We just multiply both sides by 5.
- You get y is equal to 5 sine of 15 degrees.
- So that's pretty cool.
- We know this y.
- We know the x.
- Can we figure out what this length is?
- Let me draw it in yet another color.
- Can we figure out what the length of this side is?
- Well we know from the prompt that this whole
- length is 10, right?
- We know this whole length is 10.
- And we know that this little part of it is x, which
- is, we figured out is 5 cosine of 15 degrees.
- So let's call this z.
- We know that z is equal to 10 minus x, right?
- Because this whole thing is 10.
- This thing is x.
- So z equals 10 minus x.
- And we already figured out what x is equal to.
- Z is equal to 10 minus this.
- That's what x is.
- We didn't even have to use the variable x.
- We could have just written it like that.
- So 10 minus 5 cosine of 15 degrees, right?
- That's this side.
- Z is equal to 10 minus 5 cosine of 15 degrees.
- So we know z.
- We know y.
- This is a right triangle.
- And we're looking for this yellow side, whatever
- we want to call it.
- That's the answer to the problem.
- Well this is starting to look easy.
- This is just the Pythagorean Theorem.
- So let's just-- y squared plus z squared is going to be equal
- to our-- let's just call this.
- I don't know.
- Let's call this m.
- Picking an arbitrary letter.
- Let's call that m.
- So y squared plus z squared is going to be equal to m squared.
- So we could just say-- Let's just write that down.
- I don't know why I picked m.
- Really just to confuse you, I think.
- So m squared is equal to-- Well what's y squared? y squared
- is 5 sine of 15 degrees.
- So it equals-- Let me do it in that color.
- 5 sine of 15 degrees squared plus-- What's z squared?
- z is this, right?
- z squared plus-- I'm running out of space --10 minus 5
- cosine of 15 degrees squared.
- And now I just have to simplify this.
- And if you have a calculator you could do this without
- any simplification.
- But I want to get it as simple as I can.
- So let me just-- So what's this squared?
- Well this is equal to-- Let me draw a line here because I
- don't want to get too messy.
- But I need all of this space.
- So this is equal to-- What color am I using now?
- This is equal to-- And remember, this is m squared.
- We're going to have to take the square root of all of this
- at the end to solve for m.
- This is equal to 25 sine squared 15 degrees, or sine
- of 15 degrees squared.
- That's just how you write it.
- And then we do a little bit of FOIL here to expand this.
- So plus 100 minus-- this is just expanding this expression
- --minus 100 cosine of 15 degrees plus 25 cosine
- squared of 15 degrees.
- All I is I expanded.
- I said, well this whole expression squared is equal to
- this squared minus 2 times the two things multiplied out.
- So that's 100 cosine of 15 degrees.
- And then I just squared this last term, which is plus
- 25 cosine of 15 degrees.
- If that confused you, you might want to review the
- multiplying expressions.
- So let's see if we can simplify this further.
- So if we take this term and this term, we could simplify
- that to 25 times sine squared of 15 degrees plus
- cosine squared of 15.
- And you could skip all this and just use a calculator and
- figure out this exact value.
- But I'm just going to keep working on it just because
- I like to get it as simple as possible before I
- use the calculator.
- So that term and that term are that.
- And then it's plus this stuff.
- Plus 100 minus 100 cosine of 15 degrees, right?
- And then-- What is the sine squared of 15 plus the
- cosine squared of 15?
- That's one of our most basic identities, right?
- That's the Pythagorean Identity.
- Sine squared of x plus cosine squared of x, or of
- theta, is just 1, right?
- So this term becomes 1.
- And I'm running out of time.
- So I will continue in the next video.
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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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