00:01
For vector w and quadrant 1, we want to graph it writing component form and write it in i and j hat form.
00:10
So the graph it will first make a rough sketch, so we know it's quadrant 1, so it's somewhere here.
00:18
So we'll label these sides, x, and y.
00:24
So we know that sine of theta is equal to the opposite over a hypotenuse, so y over h, but we know that h is the magnitude and theta is 74 .5.
00:42
So we have y to sine theta h, which is equal to 9 .5, sine 7 .5 degrees.
01:00
And this is approximately 9 .6.
01:07
Now for cosine, you know that cosine is equal to adjacent over hypotenuse, so x over h.
01:16
So solve it for x, we have x is equal to h times cosine theta.
01:23
So this is equal to 9 .5 cosine 74 .5...