00:01
So i've drawn the situation here.
00:03
We have a right triangle with right angle at big r, and i filled in the side lengths that they've given us.
00:09
And now we want to start solving for the unknowns here.
00:13
We first want sign of big p.
00:15
Sign is opposite over hypotenuse, and from p's perspective, the opposite is three halves.
00:22
So we get three halves over three, and so this is just equal to one half, which in decimal form is 0 .5.
00:33
Cosine of p, well, the adjacent is 3 root 3 over 2, and then we do over 3 again.
00:44
And so the 3 is going to join the 2 in the bottom and become a 6.
00:49
And when we simplify this, this is going to be root 3 over 2.
00:53
And at this point, i need to plug this into my calculator.
00:57
And when i do that, i get 0 .87.
01:02
Now tangent of p is opposite over adjacent, and the opposite is three halves, the adjacent is 3 root 3 over 2.
01:14
And so now if you flip the bottom fraction, this is going to be 3 halves times 2 over 3 or 3.
01:24
The 2s will cancel, the 3s will cancel, and we're going to be left with 1 over root 3, which in decimal form is 0 .58...