The Guild Season 3 Full Season with Trivia Annotations from Creator Felicia Day & Producer Kim Evey!

Uploaded by khanacademy on 03.11.2007

Transcript:

Now we know some trigonometry let's use that trigonometry to solve some word problems.

so let's get started

this is exciting. this is a navigation problem

so I have a ship

and it needs to go from point A to point B

so let me draw that

so I am 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.. 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 ten kilometres

ten kilometres

so this guy sets sail and because of whatever reason, he's a bad navigator or the water is flowing faster upwards, maybe the water is flowing in this direction. I'll 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 five kilometres

so he travels for five kilometres

let me do that in blue-green

he travels for five kilometres

and after five kilometres he realises that he has gone off track

and somehow, I don't know, maybe using his astrolabe, I don't really know what one is, that's what people use when they are on ships to figure out where they are I think

he realises that he is fifteen degrees off course

so he is fifteen degrees off course

so what that means is that between the path he took and the path he should have took is fifteen degrees

at least that's my interpretation of it

so this is fifteen degrees

and of course he has travelled five kilometres in the wrong direction, fifteen degrees in the wrong direction

so what he needs to know is

so this, let me draw a little ship, so this is where the ship is now, that's my ship

so now this guy, this poor captain needs to know 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 colour

how far is this distance right, here

how far is this distance?

this may just involve some trigonometry

so how can we figure out this distance?

so let's just break down what we know and that's how I do trig problems

I just keep messing around with them until I get the right answer (laughs)

so let's see if we can tackle this problem that way

so what can we figure out?

we know that this is five kilometres

this side is five kilometres

we know that this is ten kilometres

we know this angle

well one thing I know I can figure out

let's just make some right angles so that we can start using some trigonometry

so let's draw, let me draw a right angle here

right just drop it straight down

so can I figure out what this side is? right here?

let me draw that in a different color

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, and we know the angle we want to figure out the adjacent

what trig function should we use?

let me write down our soh-cah-toa

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 fifteen degrees

so I'm going to write over this just not to waste space

cosine of fifteen degrees is equal to this brown side

so let's just call this, I dunno, I'm just going to call it x, right

so this is equal to x over the hypotenuse

over five, right

and if we solve for x we get x is equal to five cosine of fifteen degrees

ah, maybe we made progress, maybe we didn't

let's keep trying

alright, now let's see if we can figure out this side of this triangle

well this is to this angle, this is the opposite, right

and we know the hypotenuse, so what trig identity should we or what trig function should we use?

well if we know the opposite.. oh, if we want to figure out the opposite and we know the hypotenuse and we know the angle

so what trig function deals with opponuse.. opposite (laughs) I was going to say hypotenuse but anyway

what trig function deals with the opposite and the hypotenuse

well that's sine right?

so the sine, if we call this y, we know that y is equal to

well doing it the same way

sine oh.. uh.. five sine of fifteen degrees

right I skipped a step, right, because we could say that sine of fifteen degrees is

sine of fifteen degrees is equal to the opposite, y, over the hypotenuse, over five

and then that step takes us here, we just multiply both sides by five and you get

y is equal to five times sine of fifteen degrees

so that's pretty cool, we know this y, we know this 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 problem that this whole length is ten

right we know this whole length is ten

and we know that this little part of it is x

which we figured out is five cosine of fifteen degrees

so let's call this z

we know that z is equal to ten minus x

right, cos this whole thing is ten, this thing is x, so it's easy to go ten minus x

and we already figured out what x is equal to

z is equal to ten minus this, that's what x is

you didn't even have to use the variable x, we could have just written it like that

ten minus five cosine of fifteen degrees

right, that's this side, z

z is equal to ten minus five cosine of fifteen 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, you know, y squared plus z squared is going to be equal to our, let's just call this

I don't know, let's call it 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 can 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

so it equals, let me do it in that color

five sine of fifteen degrees.... squared

plus what's z squared?

z is this, right

z squared

plus, I'm running out of space

ten minus five cosine of fifteen degrees... squared

and now I just have to simplify this

and if you had 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?

this is equal to... let me draw a line here 'cos I don't want it 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

so this is equal to twenty five sine squared of fifteen degrees, or the sine of fifteen degrees squared, that's just how you write it

and then we do a little bit of foil here, to expand this

plus one hundred minus

this is just expanding this expression

minus one hundred cosine of fifteen degrees

plus twenty five times cosine squared of fifteen degrees

all I did is I expanded

I said well this whole expression squared is equal to this squared

minus two times the two things multiplied out, so that's a hundred cosine of fifteen degrees

and then I just squared this last term

which is plus twenty five cosine of fifteen 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 can simplify that to

twenty five times sine squared of fifteen degrees plus cosine squared of fifteen

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 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 one hundred minus one hundred cosine of fifteen degrees

right? and then what is the sine squared of fifteen plus the cosine squared of fifteen?

that's one of our most basic identities right?

that's the Pythagorean identity

sine squared of x plus cosine squared of x, or theta, is just one

right, so this term becomes one

and I'm running out of time so I'll continue in the next video