Were on problem 61, it says the point (-3, 2) lies on a circle whose equation is (x +3)² + (y +1)² = r², which of the following must be the radius of the circle? So, the way to think about it is, is that this point satisfies this equation right. So, if we put this point in it, well you know any point on the equation will satisfy both sides of this equality signs so all we have to do is substitute the x and the y here and see what r² has to be equal too.

So, let’s do that, so we substitute the minus three for x so you get, let me do it here. You get (-3 + 3)² + (2 + 1)² = r². So -3 + 3 = 0, 0² = 0, and then (2 +1)² so 3² = r² because r² = 9 and then r = 3 because you can't have a negative radius right but we see immediately that r = 3. So, all you have to do is substitute the x and the y values right because any point that satisfies this equality is on the circle to find by this equation. They say this point is on the circle so you just have to substitute them in and just solve for r.

Problem 6, looks like we’re going to do some trigonometry. In the figure below, if sin of x = 5/13 what are the cos x and tan x? And I don’t know if you’ve seen the basic trigonometry videos, you might want too but a good mnemonic for memorizing sin, cos, tangent is Socatoa. And that means so is sin is equal to opposite over a hypotenuse, cos is equal to adjacent over hypotenuse and I play with this in a second. And tangent you might have guessed it’s equal to opposite over adjacent. So, what is that mean? What is all in this mnemonic? So, just you might want to remember Socatoa then you could break it down like that. So that means, so if I took a sign of this angle, that means the opposite side of this angle over the hypotenuse is equal to the sign of this angle so they tell us that the sin. Let’s call this the opposite. That’s the opposite side, this is the hypotenuse right. This is the adjacent side right because adjacent to the angle, this one’s opposite, hypotenuse adjacent. So, the sin of x is equal to we know from our mnemonic Socatoa is equal to opposite over hypotenuse and they tell us that that is equal to 5/13. So the ratio, so opposite over hypotenuse is equal to 5/13.

Now, we know that this is the ratio between the two so we don’t know, you know this could be 10 and this could be 26. This could be 1 and this could be 13/5, who knows, but it actually doesn’t matter because that’s what’s needed by trigonometry. It’s all about the ratios. So, let’s just assume that this is 5, that the opposite is equal to 5and the hypotenuse I equal to 13. So, if the opposite is 5 and the hypotenuse is 13 what will be the adjacent be equal to? Well, you could use it by tag and theorem. So, we could say the adjacent squared, right a² + the opposite squared so +5² + 25 = 13², 13² is 169 I think. If you subtract 25 from both sides of this equation, you get a² = 144, a = 12. So, a = 12. And we don’t know that a is definitely equal to 12 but we know that the ratio of say the opposite to adjacent is 5 to 12 because we just assumed that the opposite is 5.

Anyway so, they will know what are cos of x and tan x. Socatoa, cos of x is equal to the adjacent over the hypotenuse. The adjacent is 12, hypotenuse is 13 = 12/13. That’s the cos of x and the tan of x = opposite over adjacent (toa). So, opposite is 5, adjacent is 12. Go to 5/12 and we’ll see what choice that is, that’s choice a, cos of x is 12/13, tan x = 5/12.

Next question, oh looks like they want us to learn a lot of trigonometry in geometry which is good. This can warm you up for the trig. In the figure below sin a = 2.7 so this is called this angle A. The side of that is equal to 0.7. They say what is the length of AC? So we want to know that. Let’s call that x right. So Socatoa, So tells us that sin of sum angle, let’s call that data is equal to the opposite over the hypotenuse. So, sin A in this example. Sin A is going to be equal to the opposite 21 over the hypotenuse, over x and it tells that the sign of A is equal to 0.7 so that’s also equal to 0.7. So now, we could just solve this equation for x and we’re done. Let’s see, so if you multiply x times both sides, you get 21 = 0.7x, 21/0.7 = x. So x = 30 and that’s length AC. That’s choice C.

Next question 64, approximately, how many feet tall is the streetlight? Okay, so we can use some trigonometry here. So, let’s see we have, if we know this angle and they give us that all of the trick ratios for that angle, we’re trying to figure out the height so if I write Socatoa, what do we have, what are we trying to figure out? So, they gave us the adjacent right. This is adjacent to the angle, it’s right beside it, this is adjacent. The height that we’re trying to figure out this is the opposite. So, if we can have a trick functions that deals with the opposite and the adjacent, well that’s tangent right. TOA, tangent of any angle is equal to the opposite over the adjacent. In this case, tangent of 40 degrees is going to be equal to the opposite, the opposite is h that’s what we were trying to solve for over the adjacent. The adjacent is 20 feet. And tan of 40 degrees isn’t something that most people have memorized so that’s okay because they gave it for to us. Tangent of 40 degrees is 0.84, so this is 0.84 so we get 0.84 is equal to h/20 so we can multiply both sides of that by 20 and we get h is equal to 20 times 0.84 and that is equal to 16.8 and that is choice C.

Problem 65, right triangle ABC is pictured below, ABC, which equation gives the correct value for BC. So, this is what they want us to figure out. This is BC right there. So, let’s read them. Let me write Socatoa. I actually do this a lot. Socatoa, okay, so there is the sign, so all of this are the sign of 32 degrees is equal to BC/8.2, is that right. Sin is opposite over a hypotenuse. They are doing BC is definitely the opposite but they’re doing the opposite, 8.2 is not the hypotenuse, 10.6 is the hypotenuse. So, they are doing, this is the adjacent, so this is wrong right. This is the opposite and that’s the adjacent so this should be a tangent. Tangent of 32 = BC/8.2. This is the adjacent side. Adjacent of 32 degrees and that’s the opposite, that’s the hypotenuse so that’s not right. Choice B, cos of 32, cos is adjacent over hypotenuse. So, cos of 32 should be adjacent 8.2/10.6, so this should be in a 0.2 here. So, this isn’t right.

So, the next one tan 58 degrees, where they’re getting that 58? Well, they know that this is 32, this is 90 so this is going to be 180 - 32 - 90 because the angles and the triangle up to 180. So, this angle right here is 58 right, that’s 58 degrees. And now if we use that angle we have to relabel opposite and adjacent and all of that. So, from this angles point of view tangent is opposite over adjacent. So now the opposite side, so if we write the tan of 58 is equal to the opposite side should be equal to 8.2 over its adjacent side over BC. Alright, this is adjacent to this angle, it was opposite this angle but this is adjacent to this angle. So, that’s what they wrote. So, choice C is correct and we’re done. I’ll see you in the next video. Well, we’re not done the whole thing, I‘m done with this video. See you soon.