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- [Instructor] We're
told a boat is traveling

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at a speed of 26 kilometers
per hour in a direction

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that is a 300 degree rotation from East.

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At a certain point it encounters a current

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at a speed of 15 kilometers
per hour in a direction

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that is a 25 degree rotation from East.

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Answer two questions
about the boat's velocity

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after it meets the current.

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Alright, the first question is,

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what is the boat's speed
after it meets the current?

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And it says, round your
answer to the nearest 10th.

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You can round intermediate
values to the nearest 100th.

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And what is the direction
of the boat's velocity

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after it meets the current?

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And they say the same, well,
they actually here it say,

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round your answer to the nearest integer

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and you can round intermediate
values to the nearest 100th.

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So like always pause this video

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and see if you can work through this.

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All right, now let's
work on this together.

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So first let's visualize
each of these vectors.

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We have this vector 26 kilometers
per hour in a direction

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that is a 300 degree rotation from East.

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And we have this vector
15 kilometers per hour

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in a direction that is a 25
degree rotation from East.

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And so let me draw some axes here.

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So let's say that is my Y-axis.

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And then let's say that
this over here is my X-axis.

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And then that first vector
300 degree rotation from East,

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East is in the positive X direction.

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This would be 90 degrees,
180 degrees, 270 degrees.

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I'm going counter-clockwise
cause that's the convention

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for a positive angle.

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And then we'd go a little bit past 270,

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we would go right, right over there.

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And the magnitude of this vector
is 26 kilometers per hour.

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I'll just write a 26 right over there.

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And then this other vector
which is the current

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15 kilometers per hour in a direction

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that is a 25 degree rotation from East.

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So 25 degree rotation might
be something like this

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and it's going to be shorter.

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It's 15 kilometers per hour.

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So, it's going to be
roughly about that long.

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I'm obviously just approximating it

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and I'll just write 15
there for its magnitude.

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So we can visualize what the boat's speed

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and direction it is after
it meets the current.

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It's going to be the sum
of these two vectors.

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And so if we wanted to
sum these two vectors

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we could put the tail of one
at the head of the other.

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And so let's shift this
blue vector down here.

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So it's at the head of the red vector.

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So it would be something like this.

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And so our resulting speed
after it meets the current

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would look something like this.

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We've seen this in many
other videos so far

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but we don't want to just
figure it out visually.

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We want to actually figure
out its actual speed

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which would be the
magnitude of this vector

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and its actual direction.

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So what is the angle?

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And we could say it as a positive angle.

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So what the rotation,
the positive rotation

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from the positive X-axis or from due East.

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So to do that, what I'm
going to do is represent each

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of our original vectors in
terms of their components.

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And so this red vector up here

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and we've done this multiple
times explaining the intuition.

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It's X component is
going to be its magnitude

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26 times the cosine of this angle,

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cosine of 300 degrees.

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And it's Y component is
going to be 26 times the sine

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of 300 degrees.

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If that's unfamiliar to you,
I encourage you to review it

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in other videos where we first introduced

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the notion of components,
it comes straight out

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of our unit circle
definition of trig functions.

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And similarly, this vector right over here

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it's X component is going
to be its magnitude times

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the cosine of 25 degrees.

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And it's Y component is
going to be 15 times the sine

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of 25 degrees.

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And now when we have
it expressed this way,

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if we want to have the resulting vector,

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let's call the resulting vector S

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for maybe the resulting speed.

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Its components are going to
be the sum of each of these.

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So we can write it over here.

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Vector S is going to be equal to

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it's going to be the X
component of this red vector

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of our original speed vector.

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So, 26 cosine of 300 degrees

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plus the X component of the current.

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So, 15 times cosine of 25 degrees
and then the Y components.

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Once again, I add the
corresponding Y components

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26 sine of 300 degrees

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plus 15 sine of 25 degrees.

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And now we could use a
calculator to figure out

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what these are, to say what
these approximately are.

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So first the X component,
we're going to take

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the cosine of 300 degrees, times 26,

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plus I'll open a parenthesis here.

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We're going to take the cosine
of 25 degrees, times 15,

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close our parentheses.

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And that is equal to 26.59 if
I round to the nearest 100th.

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26.59.

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And now let's do the Y component.

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We have the sine of 300 degrees, times 26,

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plus I'll open parentheses,

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the sine of 25 degrees
times 15, close parentheses,

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is equal to negative 16.18 to
round to the nearest 100th.

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Negative 16.18.

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And let's just make sure that
this makes intuitive sense.

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So, 26.59.

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So we're going to go
forward in this direction

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26.59 on the X direction.

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And then we go negative
16.18 in the Y direction.

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So this does seem to match our intuition

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when we tried to look at this visually.

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So we now have the X and Y components

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of the resulting vector

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but that's not what they're asking for.

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They're asking for the speed
which would be the magnitude

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of this vector right over here.

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And so I could write the
magnitude of that vector

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which is going to be its speed.

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We'll just use the
Pythagorean theorem here.

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It is going to be the
square root of this squared

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plus this squared, because once again

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this forms a right triangle here.

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And we review this in other videos,

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it's going to be the square
root of 26.59 squared

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plus negative 16.18 squared
which is approximately equal to,

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they want us to round to the nearest 10th,

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26.59 squared plus.

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And it doesn't matter that
there's a negative here

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cause I'm squaring it.

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So I'll just write 16.18
squared is equal to that.

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And then we want to take
the square root of that.

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We get 31 point, if we round
to the nearest 10th, 31.1.

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So it's approximately 31.1 and
we'll write the units here,

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kilometers per hour is
the speed the boat's speed

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after it meets the current.

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And now the second question
is what is the direction

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of the boat's velocity
after it meets the current?

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Well, one way to think about it is

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if we look at this angle right over here

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which would tell us the direction

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the tangent of that angle,
theta, let me write this down.

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Tangent of that angle theta.

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We know your tangent is your change in Y

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over your change in X.

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You can even view it as the slope

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of this vector right over here.

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We know what our changes in X or Y are.

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Those are X and Y components.

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So it's going to be our change in Y

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which is negative 16.18, over 26.59,

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our change in X.

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And so to solve for theta, we could say

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that theta will be equal
to the inverse tangent.

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And we'll have to think
about this for a second

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because this might not get us
the exact theta that we want

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because the inverse tangent function

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is going to give us something
between positive 90 degrees

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and negative 90 degrees.

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But the number we want, actually it looks

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like it's going to be
between 270 and 360 degrees

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because we're doing a,

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we want to think about a positive rotation

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instead of a negative one but
let's just try to evaluate it.

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The inverse tan of this,

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of negative 16.18, over 26.59.

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16.18 negative

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divided by 26.59 is equal to this.

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And now I am going to take
the inverse tangent of that.

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And that gets us negative 31
degrees, which makes sense.

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This looks intuitive sense

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that if you were to do
a clockwise rotation

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which would be a negative
angle from the positive X-axis

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it looks like what we
drew, but let's just go

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with the convention of
everything else here.

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And let's try to have a positive angle.

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So what we can do is
add 360 degrees to that

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to make a full rotation around.

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And we essentially have
the equivalent angle.

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So let's add 360 to that to
get that right over there.

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So if we round to the nearest integer

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we're looking at
approximately 329 degrees.

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So theta is approximately 329 degrees.

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So here, when I said theta is
equal to this I could write

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theta is going to be equal
to this plus 360 degrees.

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Now what's interesting is, I
was able to add 360 degrees

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to get to the exact same place.

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If we had a situation where
our angle was actually

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this angle right over
here not the situation

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that we actually dealt with,

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but if it was in the second quadrant,

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we would have gotten this theta.

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And we would have had to
be able to realize that,

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hey we're dealing with the second quadrant

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that has the same slope.

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So instead of adding 360 degrees

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we would have added 180 degrees.

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And we've also covered that
in other videos as well.