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.