Some vectors are drawn to scale on a square grid. The green vector is the vertical component of the red vector. The blue vector is the horizontal component of the red vector. What is the angle between the red vector and its horizontal component?
Okay, looking at our square grid, we see these three vectors, the red vector, the green vector, and the blue one. And we’re told that the green and the blue are components of the red vector. Specifically, the green vector represents the vertical component of the red, and the blue vector represents its horizontal component.
Along with the square grid and these three vectors, we see a protractor aligned so that 90 degrees on the protractor aligns with the green vertical component vector and zero degrees on the protractor aligns with the blue vector. And moreover, the tail of the red vector is positioned right where those vertical and horizontal lines intersect. This means our protractor is in the perfect position to help us answer this question. What is the angle between the red vector and it’s horizontal component, that is the blue vector?
If we were to sketch this angle in on our grid, it would look like this, starting at the red vector and going to the blue vector, the horizontal component of the red vector. We could give this angle a name. We could call it 𝜃. And it’s the 𝜃 that we want to solve for. The way we’ll do it is by reading this angular distance on our protractor. Now, there are a couple of different ways to do this.
Notice, for example, that on the outside of our protractor, the angles go from zero degrees on the left all the way up to 180 degrees on the right. While on what we could call the inner layer, they go the opposite direction, starting at 180 ending at zero. The reason for these scales going in opposite directions is to make it straightforward to always measure a positive angle with our protractor. And indeed, the angle we’ll measure, the angle we’ve called 𝜃, will be a positive value.
One way to measure it is to start at zero degrees on the right-hand side of our protractor and move up from there until we reach the red vector line. Notice, though, that on this interior scale of angles, the one we’ve indicated in pink, the finest resolution of angles is 10 degrees. That is, it goes from 180 to 170 to 160 degrees to 150 to 140 and so on. On the other hand, the outermost scale, the one we’re indicating in orange, has a resolution down to one degree. We can see this by counting the tick marks on that scale.
Say we start at 50 degrees right here, then we count one, two, three, four, five tick marks to the halfway point, and then continuing on six, seven, eight, nine, 10 tick marks to get up to 60 degrees from 50. This confirms that the angular distance between adjacent tick marks is one degree. And that tells us if we measure using this outermost scale, we can record the value of 𝜃 to the nearest degree. So, then, let’s use this outer scale to measure 𝜃, and here’s how we’ll do it.
We can see the red vector crosses the outermost angle scale right here. And if we count tick marks, that’s one tick mark, two tick marks away from the angle marked out as 130 degrees. Since those two tick marks are in the direction of a smaller degree measure, in the direction of 120 degrees. That means that if we start from zero degrees on the outermost scale and then work our way all the way around to the angle that we’ve just measured, that angular measure will be 130 minus two degrees, or 128 degrees.
But we can see that that’s not the value of 𝜃. 𝜃, rather, is the difference between the angle we’ve just measured and 180 degrees. That’s because if we started at zero degrees on our outermost angle scale and worked all the way around until we reached the blue horizontal component of the red vector, we would’ve crossed through 180 degrees. So, to find 𝜃, to find the angle between the red vector and its horizontal component, we’ll subtract 128 degrees from 180 degrees. That subtraction will give us this angular distance right here, the one we’ve called 𝜃, what we want to solve for.
180 degrees minus 128 degrees is 52 degrees. That’s our answer. And if we look back to the innermost angle scale, the one that goes in the reverse direction of the outer angle scale. We see that this supports our result that the red vector looks to be just past 50 degrees on that angle scale. Our final answer, then, is that the angle between the red vector and its horizontal component is 52 degrees.