0:00
Hello.
00:01
So here we want to find all three angles of our triangle having vertices at, well, the origin, 0 .0.
00:07
Point a here to point 27 and b, which is the point 6 comma 3.
00:11
So then we let the vector here, oa, be equal to our vector u, which is going to be equal to a minus zero.
00:20
So that's going to be 27.
00:22
And then we let ob be equal to our vector v.
00:30
Which is going to be equal to b minus 0, so that's going to be 6 .3.
00:36
Then we have, let's say, angle a, o, or angle a, o, b, call that theta, maybe theta 1.
00:49
Then we have here that cosine of theta 1 is going to be equal to uv over the magnitude of u to the v.
00:59
So we get 2 .7 times 6 .3.
01:03
That's going to give us a 12 plus 21, which is going to be 33, and divided by the square root of 53 times a square root of 45, just a square root of 45 times 53, which is going to give us that theta 1 then is going to be equal to the inverse cosine, this here of 33 over the square root of whatever 45 times 53 and we get then that theta 1 is going to be equal to about 47 .47 degrees.
01:53
Okay.
01:54
And then next we have that b -o is going to be equal to well negative o b.
02:05
So that's going to be equal to negative v, which is going to be equal to negative 6, negative 3...