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Direction of vectors from components: 1st & 2nd quadrants

Sal first finds the direction angle of a vector in the first quadrant, then moves onto a trickier one in the second quadrant.

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  • leafers tree style avatar for user Garner
    At Sal simply presses the inverse tangent button on the calculator and is given a new number (which I assume to be degrees). I don't know what this button is doing or how it does it. I've tried looking through Khan about Tangent and there is a lot of videos. Can any one point me to a basic explanation of what this key does and how it does it?
    (15 votes)
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  • leaf blue style avatar for user mikagussler
    Are there any videos that help or teach us how to find a vector using its magnitude and an angle θ ? For example: Find the vector V given the magnitude and the angle it makes with the positive x-axis.

    ||v||= 8 θ=270°
    (8 votes)
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  • aqualine seedling style avatar for user Hadrien Lacroix
    Regarding the second problem starting at , I solved it in a different way:
    atan(5/6) for the left side of the angle angle in blue, and add 90 for the right side of the angle in blue, which gives you 129.8056 as well.

    Is there any particular reason as to why we should use one method instead of the other? I understand the y/x method seems more straightforward and systematic but it gets me confused with the tan = opposite / adjacent formula, since here y/x is 6/-5 and the tan formula gives us 5/6.
    (5 votes)
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    • female robot grace style avatar for user Timon
      As long as you are carefull your method is perfectly fine. However, some people might make the mistake of doing atan (6/5) which gives you the angle below the vector w in this video and the negative x-axis.
      If you do (180 - atan (6/5)) you would still be good though.
      (7 votes)
  • hopper cool style avatar for user Qazim Dervishi
    Does the change in the x direction always has to be i^ and change in y to be j^ ? Or you can use other letters to denote these unit vectors ?
    (4 votes)
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  • orange juice squid orange style avatar for user Ines
    Can someone explain to me why tan^-1 of (-6/5) gives an answer in he 4th quadrant, and not in the 2nd? Sal said something about calculators at giving only answers from -90 to 90 degrees but I still don't get it. And why 180 degrees + the answer?
    (4 votes)
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    • mr pink red style avatar for user andrewp18
      If you take a look at the tangent function's graph, you can see that it clearly does not pass the horizontal line test, meaning it is not invertible. However, to invert it, we restrict its domain to -90˚ to 90˚. This implies that the range of the arctangent function is also -90˚ to 90˚. Hence, all angles that are outputted by the arctangent function will be in quadrants IV and I. But for every angle with a certain tangent in quadrant IV, there is another angle with the same tangent in quadrant II (and likewise for quadrants I and III). To find that angle, we are essentially continuing the terminal side of our angle into the opposite quadrant and drawing a diameter of the unit circle instead of just a radius. This is equivalent to simply adding 180˚ to get the other side of the segment. This is why we add 180˚ to find the quadrants I and III solutions to the calculator's answers which will always be in quadrants I and IV. Comment if you have questions.
      (5 votes)
  • leaf red style avatar for user layaz7717
    Why is tan theta -6/5? I just don't see how that angle fits into a right triangle, and resultingly, how we can use tangent to solve for it?
    (3 votes)
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    • female robot grace style avatar for user loumast17
      Technically once the theta angle gets to 90 degrees you can't think of it as a standard right triangle.

      For instance sin(90) is tricky to say what exactly the opposite or hypotenuse would be. This means the visualization only works for when both the opposite and adjacent sides are greater than 0. or in other words we are in quadrat 1.

      Trig functions were designed to use the whole unit circle/ graph. When you leave the first quadrant it is more accurate to say the "opposite" side is the y coordinate and the "adjacent" side is the x coordinate. As long as the angle you are measuring is at the center of the plane as shown in the video.

      Does that make sense to you?
      (1 vote)
  • blobby green style avatar for user atty225
    at , why do you take the absolute value here and not in previous calculation... s i worked out that x would still be arctan(-6/5)... since the x component is -5?
    (2 votes)
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    • leafers seedling style avatar for user CCDM
      Since he is working out the acute angle x, relative to the negative x axis, you can work with magnitudes. You can then get theta as 180 (positive counter clockwise rotation to the negative x axis) and then a -50 degree clockwise negative rotation back to angle of interest.

      OR, you could have done the tan(theta) = -6/5 but then you get the negative angle -50 and you have to remember that tan has a k(pi) repetition (where k is +/- integer) and then figure out that k must be 1 in this problem to put you back in the desired Quadrant II
      Either way, you get the same answer.
      (1 vote)
  • leaf blue style avatar for user Jessica Ezemba
    Hello,
    In the video, Sal said that most calculators give you an arc tangent from the range of 90 degrees to -90 degrees. Can someone please tell me what range calculators give when solving for arc sin and arc cosine? Thank you!
    (1 vote)
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    • winston baby style avatar for user Broken Fixer
      arcsin has a range of -90 to 90. arccos has a range of 0 to 180. (If you graph x = sin(y) in desmos, you see that the graph curves back on itself, failing the vertical line test. Typing y = arcsin(x) will give the same graph, but restricted to a set where the output is single-valued. Make sure that your calculator is set to degrees rather than radians - the range in radians is -pi/2 to +pi/2.)
      (1 vote)
  • blobby green style avatar for user Abhishek Dwivedi
    How to find the inverse of trigonometry functions without Calculator ?
    (1 vote)
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  • starky ultimate style avatar for user mnidadavolu
    For the second question, could you go positive 50 degrees from the negative part of the x-axis?
    (1 vote)
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    • leafers seedling style avatar for user CCDM
      Nope. Remember the definition of positive and negative angles. Positive angles rotate counter clockwise (CCW) and negative angles rotate clockwise (CW) from the positive x axis. If you went positive 50 degrees from the negative part of the x axis then you would have 180 degrees (moving from positive x axis to negative x axis) + another 50 degrees would put you down in the Quadrant III which is NOT where you want to be.

      There are many ways to think of these things. How I would have thought about it is, like Sal, find the angle theta relative to negative x axis (since you have the opposite/adjacent values). Then I would have rotated +180 degrees CCW from positive x axis to negative x axis and then rotate CW -50 degrees to get back to the desired Quadrant II location. 180-50 degrees gives proper answer (ok, I rounded the theta but you got that)
      (1 vote)

Video transcript

- [Voiceover] So what we're going to do in this video is look at a series of vectors, and we're gonna draw them in standard form, where their initial points or their tails are gonna sit at the origin. And we're gonna see if we can figure out the angles that they form, the positive angles that they form with the positive x-axis. And like always, pause this video and see if you could figure out what these thetas are going to be on your own, and we're gonna figure 'em out in degrees. So in this first one, I have the vector u. If we were to write it in unit-vector notation, sometimes this is called engineering notation, we would say it's three times the unit vector in the horizontal direction, the i unit vector, plus four times the unit vector in the vertical direction or you could view this as the x-component is three and the y-component is four, and you see that here. If you start at the origin, we're gonna move three in the horizontal direction, and we're gonna move four in the vertical direction. Now to figure out this theta, there's a couple of ways to think about it. We could just construct a right triangle for this one in particular. We could say, okay, the x-coordinate is three. So if we were to create a right triangle here, this side would have a length of three, and then we have a height of four or the y-coordinate is four. So this side has a length of four, and we know just from even our basic SOHCAHTOA definitions of trig functions that what trig function involves the opposite of an angle? So the opposite of an angle, and the adjacent of an angle? Well, tangent does. SOHCAHTOA. So we know, we know that the tangent of theta is going to be equal to the length of the opposite side, which is four, over the length of the adjacent side, over three. And so if we wanted to solve for theta, we could just say that theta is equal to the inverse tangent. Sometimes people say arc tangent of 4/3, of four over ... four over three, and let's evaluate this. And I'm gonna get my calculator out to do it. So I wanna take four divided by three, which is that, and I had already had pressed this button right over here that takes my tangent and makes it into an inverse tangent. So I'm gonna take the inverse tangent of this, and I get, I get roughly 53.1 degrees. So this is approximately 53.1 degrees, which looks about right. This looks like a little bit more, even though I didn't draw it completely super precisely. I hand-drew this. It looks a little bit more than a 45 degree angle, so that feels, that feels pretty good. Now another thing you might have said, "Hey. Well, look. "Four is the y-component. "Three is the x-component, and so maybe tangent "of theta is always going to be "the y-component over the x-component." And that, in fact, is the case. And that actually comes straight out of the unit circle definition of the trig functions, where you could say that if you have a unit circle, if you have, I could draw it right over here. So let me draw the coordinates. So if I were to draw a unit circle right over here, and if I were to have some line, here we're thinking about vectors, that the angle formed with the positive x-axis, the tangent of that angle, tangent of theta is going to be the y-coordinate, where we intersect the circle, over the x-coordinate. And so you could imagine if you made a unit circle right here, if you made a unit circle right over here, the ratio between the y-coordinate and the x-coordinate of this point right over here where we intersect the circle is going to be the same thing as the ratio between four and three. So this is going to be, this also is going to be one and 1/3 or 4/3. So either way, you could think of, when you think of vectors, the tangent of the angle that it forms with the positive x-axis is going to be, is going to be the y-component over the x-component. So it's nice when everything kinda fits together like that. So let's leverage that to figure out this, to figure out this angle right over there. Well, we could say that tangent of theta is going to be equal to the y-component. The y-component over the x-component. Over, do this in a new color or let's see, over negative five. So it's gonna be negative 6/5. Or we could say theta. So let's be a little bit careful. So we could say, we could say that theta is equal to, is equal to the inverse tangent. Let me do that same color. Inverse tangent of, I'll just write it like this, negative 6/5, but I'm going to put a little question mark here to see if we feel good about the answer that you get when we do this. So let's do that. So if we do, if we do six divided by five, which'll equal to that, and then we want to make this negative. So that's negative 6/5, and we're gonna take the inverse tangents. I already pressed this button, so this is going to be an inverse tangent, not tangent. I get negative, roughly negative 50.2 degrees. So this is approximately negative 50.2 degrees. Well, does that look right? Well, no. Theta looks like it's over 90 degrees. Negative 50.2 degrees, negative 50.2 degrees is actually giving us this angle right over here. So that's giving us, that's specifying another vector, if you think about a line, another line that would have the same tangent. The same tangent. So it's really important to visualize this and think about this, and the reason why it is is because the arc tangent or the inverse tangent functions in calculators, they will give you an angle that is between negative 90 degrees and positive 90 degrees, so something that's in the fourth or first quadrant. While here, we have something that's in the second quadrant. So we have to make sure that we're thinking about it right so that we could make the appropriate adjustment. So that would give the case if we were looking at a line like that or a vector like that. So in order to figure out what, in order to figure out the actual angle, what we want to do is add 180 degrees to it. So to get the vector that goes in the other direction. So you want to add, so theta is going to be equal to or I could say approximately equal to negative 50.2 degrees plus 180 degrees. So let's do that. So let me plus 180 is equal to approximately 129.8 degrees. So theta is approximately equal to 129.8 degrees. Did I get that right? I have a very short memory. Okay, yeah; that's right. 129.8 degrees, and that looks much better. That looks clearly is an angle we're gonna go. That take us to 90 degrees, and then we're going above 90 degrees. Now another way you could have thought about this is what you know from SOHCATOA and just right triangles. We could construct a right triangle where, we construct a right triangle using some colors. Well, we know, so I could maybe just draw it like this. So this is the height of it, and what is that height? Well, the height is going to be six, and then the base, the base right over here, what is that length going to be? Well, we know that we're going from the origin, we're going five back, but if we think about just the absolute value, the length of that line, that is going to be five, and so we could figure out just using right triangles this angle. Maybe we call that x, and so we could say that the tangent of x, tangent of x is equal to six over five, opposite over adjacent. Six over five. Or that x, or you could say x is equal to the inverse tangent of 6/5, and what is that going to be? So, if we take six divided by five, 1.2. And then we're going to take the inverse tangent of that, and we get approximately 50.2 degrees. So x is approximately 50.2 degrees, which looks right. But remember, we're not trying to figure out x, we're trying to figure out, we are trying to figure out what theta, what theta is, and you can see that x and theta are supplementary. So theta plus x is going to be 180 or theta's gonna be 180 minus that. So 180 minus that. So let's just put a negative sign in front of that and add 180, and that might look familiar. So that gets to us to 129, roughly 129.8 degrees is what exactly what we got before.