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## Precalculus

### Course: Precalculus>Unit 6

Lesson 6: Direction of vectors

# 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.

## Want to join the conversation?

• 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? • 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° • 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. • 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 ? • 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? • Around , Sal talks about SOHCAHTOA. I don't really know what this stands. Can someone please explain it?
(1 vote) • 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? • 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)
• 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? • 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)
• 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) • 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)
• For the second question, could you go positive 50 degrees from the negative part of the x-axis?
(1 vote) • 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)