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## Precalculus

### Course: Precalculus > Unit 6

Lesson 6: Direction of vectors# Direction of vectors from components: 3rd & 4th quadrants

Sal finds the direction angle of a vector in the third quadrant and a vector in the fourth quadrant.

## Want to join the conversation?

- How do we get tan to the power -1?(10 votes)
- tan to the power of -1 is NOT the same as 1/tan. Please refer to

https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/trig-solve-for-an-angle/a/inverse-trig-functions-intro(17 votes)

- So the basic rule of this and the previous video is:

In Quad 1: +0

In Quad 2: +180

In Quad 3: +180

In Quad 4: +360

??(10 votes)- You are correct, But instead of blindly learning such rules, I would suggest understanding why you do that to fully understand the concept and have less confusion.

By the videos, it can easily be understood why it is so.(10 votes)

- Is there any way to find out the inverse tangent, sine, and cosine by hand?

Are there any methods?(6 votes)- Yes, but the math is too advanced for this level of study.

For example, here is the formula for the inverse sine of x (using radians, not degrees):

sin⁻¹ x = − i * ln [i x+√(1-x²)]

You will not be expected to do this kind of math, but you will be expected to memorize the inverse functions of the special angles.(14 votes)

- In Quadrant 3, is it possible to find the angle inside the triangle, and then subtract it from 270? Will that method also work?(7 votes)
- No, you can't... when dealing with angle operations along the y-axis (90,270) you convert the sign to its complementary: sin <|> cos, tan <|> cot, but when you perform operations along the x-axis (180,360) you just change the sign, preserve the function type... I recommend you watching Trigonometry videos for further explanation... it all comes out of similarity... I hope this helps if you haven't figured it out by now :)(4 votes)

- At0:25, what is the point of writing the vector as (-2i - 4j)? Because writing it as (-2, -4) is the same thing, except without the useless letters...?(4 votes)
- Why write a number such as 345 as 3.45 x 10²? The latter is scientific notation - it has its place.

Why write a vector, such as (2, 4) as 2i + 4j? The latter is engineering notation - it has its place.

Some conventions may seem pointless to you now, but if you ever get into the areas they are used, they will make total sense.(9 votes)

- In the 3rd qudrant, I did tan(270-theta) = 4/2

=> theta = 206.56 degrees. This answer isn't the same as Sal who calculates it as 243.4 degrees. Can somebody help me here?(2 votes)- tan¯¹ (4/2) =63.43°, which is in the first quadrant. To find the third quadrant angle of the same tangent, add 180°. 63.43 + 180 = 243.4°(2 votes)

- In the 'Direction of vectors' videos we are only dealing in two dimensions, so it is easy to visualise. However, with three dimensions or higher we might not be able to determine whether the tan result is correct by visual inspection. Will the rules of adding 180 and 360 still hold at these higher dimensions? Do we apply the same thinking at higher dimensions or rely on something else entirely?(2 votes)
- Why do we need exactly positive angle?

If tangent is defined at -pi/2 < x < pi/2 I feel that answer -56 degrees is correct for 4th quadrant.(2 votes) - Why in 2nd & 3rd quadrant, we add 180 degrees to the angle? and why in 4th quadrant, we add 360 degrees?

How do we know that when we should add 180 and 360 degrees to get the correct angle of the vector?

Somebody pls clarify it:((1 vote)- Taking the inverse tangent of the ratio of sides of a right triangle will only give results from -90 to 90, so you need to know how to manipulate the answer, because we want the answer to be anywhere from 0 to 360.

if both coordinates are positive, you are fine, you will get the right answer.

if both are negative, so in quadrant 3, you are taking the inverse tangent of a fraction with a negative numerator and denominator so it would be positive. If you try a vector like 2i + 3j and then -2i - 3j, you'll get the same answer. So you need to realize the tangent and angle is the same as the tangent of 180 plus that angle.

Now, if one is positive and one is negative that puts it in either quadrant 2 or 4. In both cases you are taking the inverse tangent of of a negative number, which gives you some value between -90 and 0 degrees. So for all positive ratios you take the inverse tangent of the result is between 0 and 90.

Now, if you have a positive x value and negative y value, so quadrant 4, the answer is technicallyc correct. Most answers want the value between 0 and 360, so you need one more full revolution to get it there.

Quadrant 2 meanwhile has the same logic as quadrant 3 from before.

If it helps lets use the coordinates 2i + 3j again. This makes a triangle in quadrant 1. if you used -2i + 3j it makes the same triangle in quadrant 2. The relevant angle is obviously 180 minus that angle, I will call x. Using tangent you get -x so you add 180, which is the same as 180 - x

-2i - 3j makes the same triangle in quadrant 3 where the relevant angle is 180 + x

So that means if you take the tangent of a vector in quadrant 2 or 3 you add 180 to that.

If you have -2i - 3j then you have the same triangle in quadrant 4. You could look at the relevant angle as -x or 360 - x, the 360 - x is more useful. Taking the inverse tangent gets you -x again, so adding 360 to it puts it at the appropriate range of numbers.

I really really hope that helped, if not though let me know.(3 votes)

- can anyone tell me the inverse trig values of special angles?(1 vote)
- you can't do an inverse trig function of an angle, it has to be a ratio.(2 votes)

## Video transcript

- [Voiceover] Let's get some more practice finding the angle, in these
cases the positive angle, between the positive X axis and a vector drawn in standard form
where it's initial point, or it's tail, is sitting at the origin. Here for vector A we can write
it in two different ways. In engineering notation it would be -2 times a unit vector I, that's the unit vector in the X direction, minus four times the unit
vector in the Y direction, or we could just say
it's X component is -2, it's Y component is -4. And we see that here. If we're starting at the
origin we go two to the left and we go four down to
get to the terminal point or the head of the vector. Also figure out what theta is. What we've seen before when
we're thinking about vectors drawn in standard form,
we could say the tangent of this angle is going to
be equal to the Y component over the X component. So the Y component is -4 and the X component is -2. And in the previous video
we explained why this is, it really comes straight out
of the unit circle definition of trig functions, tangent of theta is equal to the Y coordinate
over the X coordinate of where a line that
defines an angle intersects the unit circle. And I encourage you to watch that video if that doesn't make much sense. But so we could say tangent
of theta is equal to two. And so we might want to say,
if we want to solve for theta, we could say theta is equal to the inverse tangent function of two. And I'm gonna put a question mark, and I think you might know why I'm putting that question mark. So let's see what that gets us. So if we were to take two, and I wanna take the inverse
tangent not just the tangent. I wanna figure out what angle
gives me a tangent of two. So inverse tangent, it's about 63.4 degrees roughly. So this gives me theta is approximately 63.4 degrees. And I'm gonna put a question mark here. And I think you might sense why that is. If you don't, pause the video and think about why am I
putting a question mark here? Why does this angle look fishy? Well, it looks fishy because
an angle of 63.4 degrees would put us squarely in the first quadrant. If our vector looked like this, let me see if I can draw it. If our vector looked like this, so if our vector's
components were positive two and positive four then that
looks like a 63-degree angle. This looks like a 63-degree angle. But we're not in the first quadrant. Our vector A that we care
about is in the third quadrant. So it's clear that it's in
the exact opposite direction, and I think you see why. When we take the inverse tangent
function on our calculator it assumes that the angle
is between -90 degrees and positive 90 degrees. Well, here we have an angle
that's over 180 degrees. It's between 180 and 270 degrees. And so to find this angle, and this is why if you're
ever using the inverse tangent function on your calculator
it's very, very important, whether you're doing
vectors or anything else, to think about where does
your angle actually sit? What quadrant does it actually put you in because you might have
to adjust those figures. So, it's not going to be 63.4
degrees it's going to be that plus another 180 degrees to
go all the way over here. So, theta is going to be 180,
and I should say approximately 'cause I still rounded,
180 plus 63.4 degrees is going to be 200 and, what is that? 180 plus 60 is 240, so 243.4 degrees. So let's do one more. So here I have a vector
sitting in the fourth quadrant like we just did. Pause the video and see
if you can figure out the positive angle that it
forms with the positive X axis. Well, we could do the same drill and maybe we could skip a few steps here now that we've done it many times. We might wanna say that
the inverse tangent of, let me write it this way,
we might want to write, I'll do the same color. We might wanna say that
theta is equal to the inverse tangent of my Y component
over my X component of -6 over four, and we know what that is but let me just actually
not skip too many steps. So the inverse tangent of -1.5. - 1, -1.5 and once again, I get to get my calculator out and so 1.5 negative, and I wanna find the
inverse tangent of it, I get roughly -56.3 degrees. So this is approximately equal to - 53.6 degrees. Did I do that right, 56.3, 56.3 degrees. And once again, I'm gonna
put the question marks here. And why did I do that? Because the angle that it's giving, and this isn't wrong
actually in this case, it's just not giving
us the positive angle. Because if you start the positive X axis and you were to go clockwise, well now your angle is
going to be negative, and that is -56.3 degrees. But we wanna figure out the
positive angle right over here. So, there's a couple of ways that you could think about doing it. One way to think about it is well to go from this negative angle to the positive version of it we have to go completely around once. So we have to add 360 degrees. So if it's really approximately -56.3 degrees plus 360 degrees, which is going to be, what is that? So it's going to be, so it's going to be approximately, see if I subtracted 50 degrees
I would get to 310 degrees, I subtract another six
degrees, so it's 304 degrees, and then .3, so 303.7 degrees. Did I do that right? Let's see, if I add this .3 to the seven, that's gonna get to
304, then at 310 to 360. So there ya go. That is our positive angle that we form. So always really think about
what they're asking from you, or what a question is asking from you. Anyway, you get the idea.