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We've got triangle ABC here. It looks like a right triangle. We can verify that it is because
it satisfies the Pythagorean theorem. 8 squared is 64 plus
15 squared is 225. 64 plus 225 is 289,
which is 17 squared. So it satisfies the
Pythagorean theorem. So we know that this right
over here is a right triangle. But what they're
asking us is-- what is the cosine of
angle ABC, which is this angle right over
here plus 60 degrees. Now, just like
this, I don't have any clear way of evaluating it. But we do have some trig
identities in our toolkit that we could use to
express this in a way that we might be
able to evaluate it. In particular, we know
that the cosine of A plus B is equal to the cosine of
A times the cosine of B minus the sine of A
times the sine of B. So we could do the
exact same thing here with the cosine of angle
ABC plus 60 degrees. This is going to be equal
to-- and let me just write out the whole thing--
the cosine of angle ABC times the cosine of 60
degrees minus the sine of angle ABC times the sine
of 60 degrees. So this right over
here is angle ABC. This is angle ABC. And this is 60 degrees,
and this is 60 degrees. So let's evaluate
each of these things. What is the cosine of angle
ABC going to be equal to? I'll do that over here. The cosine of angle ABC. Well, we could go
back to Sohcahtoa. Let me write it down. Cosine is defined as the
adjacent-- for this angle-- the adjacent over
the hypotenuse. So cosine of angle ABC
is equal to 15 over 17. So this right over
here is 15 over 17. What is the cosine
of 60 degrees? Well, for there, we have to turn
back to our knowledge of 30, 60, 90 triangles. So if I have a triangle
like this-- so let me do my best
here-- to construct a 30, 60, 90 right triangle. So that's a 60-degree side. This is a 30-degree side. We know-- and we've seen
this multiple times-- if the hypotenuse has
length 1, the side opposite the 30-degree
side is one half. And then, the side
opposite the 60-degree side is square root of 3 times this. So it's square root of 3 over 2. So the cosine of 60
degrees-- if you're looking at this side
right over here. And let me write it in a
color I haven't used yet. So I care about the
cosine of 60 degrees. The cosine of 60
degrees is going to be equal to, once again,
adjacent over hypotenuse. One half over 1. it's going to be equal
to one half over 1, which is equal to one half. Now, let's think about the sine. Now, let's think about
the sine of angle ABC. So that's this one
right over here. So we have our
triangle right here. Sine is opposite
over hypotenuse. So opposite has length 8. Over hypotenuse, is 17. So it's equal to 8 over 17. And then, finally,
we need to figure out what the sign of 60 degrees is. Well, for the 60-degree
side of this right triangle, opposite over hypotenuse. So square root of
3 over 2 over 1, which is just the
square root of 3 over 2. So we have all of
the information we need to evaluate it. So this was the
sine of 60 degrees. This whole thing is
going to evaluate to cosine of angle ABC is
15 over 17 times cosine of 60 degrees is one half. So times one half. And then, we're
going to subtract sine of ABC, which is 8 over 17. And then, times
sine of 60, which is square root of 3 over 2. So times the square
root of 3 over 2. And now, we just have to
simplify it a little bit. So this is going to be
equal to-- if I multiply one half times this, let's see. It's going to be 15 over
34 minus-- and let's see. We're dividing by 2. So it's 4/17ths. I'll write this 4 square
roots of 3 over 17. And that's about as
simple as I could do it. If I wanted, I could have a
common denominator here of 34. And then, I could add the
2, so it could be 8 squared. So 3 over 34, but that still
won't simplify it that much. So this is a
reasonable good answer for what they are asking for. 15 over 34 minus 4 square
roots of 3 over 17.