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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.