Trigonometry word problems (part 1)

Trigonometry word problems (part 1)

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 squared is five sine of fifteen degrees
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