We are on 76. In the figure below, line AB is tangent to circle O at point A. Fair enough. Point A secant BD intersects circle O at point C and D. Right, this is secant BD. Secant just means that it intersects the point at two points. It’s not tangent. A tangent intersects a circle at exactly one point. All right? Secant BD intersects a circle at point C and D. Measure of arc AC is equal to 70 degrees. So, the measure of arc AC, this is 70 degrees, okay. And the measure of arc CD is equal to 110 degrees. That’s equal to 110 degrees. So what do they want us to figure out?

What is the measure of ABC? So, what is ABC? So what is this angle? That’s what they want us to figure out. They want us to figure out this right here – that angle. Let’s see what we can do here. So, there’s a couple of ways to think about it. Well, there’s one way if you look at the – this is 70. This is 10, right? So if you add them, this whole arc going for all the way around, that’s actually what? It’s 70 plus 110. That’s 180 degrees. So it’s actually, perfectly half way around the circle. So actually, from A to D is actually a diameter of the circle. How do I know that? – Because it goes all the way around the circle. It’s 180 degrees that the arc length is 180 degrees. So we can draw – let me draw a diagonal line. I don’t know if this thing is drawn to scale.

So A to D is actually a diameter of the circle. We know that because it’s 180 degrees – a combined arc length right there. And we know that this line is tangent at point A. This line is tangent at point A. So that has to be a right angle, right? When you’re tangent at point A, that means you’re tangent to the radius at that point. Fair enough.

We know that this arc length is 70. Is there any way to figure out what this angle right here is, because if you know this angle, we know this is 90 then we could figure out that angle. And this is something that you should just learn about, what do you call them? Inscribed angles in a circle. This is an inscribed angle because the vertex touches on one of the sides of the circle. So, an inscribed angle – this is something good to memorize. And you could play around to get a little more intuition about is equal to half of the arc length that it intersects. So, this inscribed angle intersects an arc length of 70 degrees, right? So, this is 35 degrees. It’s 35 degrees. And that’s just something good to know that this is going to be half of whatever this arc length is, right?

Now, we can figure out X because X plus 35 plus 90 is equal to 180. So, let’s see. You get X plus 35 is equal to 90. X is equal to what? X is equal to 90 minus 35 is 55, which is choice C. Now, there’s another way you can do this and this is another interesting thing about lines that intersect in circles is that, this angle measure right here, if it’s sitting outside the circle and I think they may want to be sitting inside the circle later on. But if it’s sitting outside of the circle, it’s equal to one half the difference of the arc measure that it intersects.

So example, it intersects this arc measure of 70 and it also has this arc measure right here, right? When it hits the circle and intersects at 70 degrees and when it exits the circle, it has this arc measure. This arc measure, we already figured it out is 180 degrees because that’s 180 degrees. So you know that this angle right here, you can say that X is equal to one half the difference of this two arc measures and that. We can do that because it’s outside of the circle. If X was inside the circle, then X would be one half the sum of the two measures.

So, what’s the difference? It’s 180 minus 70. So that’s equal to one half times 110. So it’s that is equal to 55, again. So, it’s nice to see that math works around – works correctly no matter how you do the problem.

Problem 77. In the circle shown below, the measure of arc PR is 140 degrees. Sure enough. The measure of angle RPQ is 50 degrees. Okay, that’s 50 degrees. What is the measure of arc PQ? What is this arc, right there?

So, we could use what we learned in the last problem. That 50 degrees is an inscribed angle, right? This is an inscribed angle. So, it’s going to be half of the arc length – half of the arc length measured in degrees or arc angle, I guess you can say, of the arc that it intercepts. So, if this 50 then this arc measure is going to be 100 degrees, right? So, what we’re trying to figure out, this arc measure right here it’s really what you have left over, right? The whole circle has 3 – let’s call this X. So, we know that X plus 100 plus 140 – well, that goes all the way around the circle, right? That’s going to be equal to 360 degrees. See, if X plus 240 is equal to 360 degrees, subtract 240 from both sides. You get X is equal to 120 degrees. Choice D.

Problem 78. The vertices of triangle ABC are, okay A is 2,1; B is 3,4; and C is 1,3. If triangle ABC is translated one unit down, so, one unit down means we’re subtracting 1 from the y’s, right? And three units to the left. So, that means you’re subtracting3 from the x’s. To create triangle DEF, what are the coordinates of the vertices of DEF? So, let’s see, we have A – A was 2,1 and that’s going to be translated to – let me write it neater. A is 2,1 and that’s going to be translated to point D. Or what did say? One unit down. One unit down means y decreases by 1. So the y coordinates is going to be – decreased that by 1, you get zero. And 3 units to the left, so that means you decrease the x coordinate by 3. So 2 minus 3 is negative 1. Actually, just looking at the choices, we’re already done, right? This is – D is the only choice that had a negative 1,0.

But let’s take a look at the other ones. Maybe somehow, they change the lettering or something. Let’s see point B was 3,4. So, what do we do with it? X, we said 3 units to the left. So, that becomes D –that becomes point E. So you take 3 units to the left. You’re , subtracting 3 from there so, that becomes a 0. If you take the y coordinate 1 unit down, that becomes a 3. So E is 0,3. So, that’s still consistent with this. So, let’s just make sure that the F works. So, C was 1,3. If we take 3 away, right? – We shifted to the left by 3. So that’s taking the 3 away from the x coordinate. So, that’s minus 2. And if then you’re shifting the y coordinate down 1, that’s taking 1 from this. So, that’s 2. So, F is minus 2,2. And so, D was definitely the answer.

Problem 79. If triangle ABC is rotated 180 degrees about the origin, what are the coordinates of A prime? So, we’re going to rotate this thing 180 degrees. And essentially, we could worry about all the coordinates, but we just want to know where A prime is. So, this is going to have – so wherever A sits after we have rotated it, right? So, let’s think about it a little bit. Let’s think about it a little bit. When we’ve rotated the C – this is going to be the new C. It’s going to be right there. And then the new A – So, let’s see. The B is going to be – when you rotate it, I just want to make sure I’m doing this right. This is almost like a – this is a visualization problem. If you go 4 – no. If you go, this is 4 to the right and up 3, right? So, if you were to rotate it around, let’s say this little arrow I drew with a 4 – If I were to rotate that 180 degrees, I just want to make sure I do it right, then that’s going to be 4 this way, right. That’s going to be 1, 2, 3 – it’s going to be 4 this way. So, you’re going to be at that point. And then you’re going to go – instead of 3 up, you’re going 3 down. So, it’s going to go 3 down. So, that’s where the A prime is going to be right at that point, right there. And that’s the point minus 5, minus 4. That’s minus 5, minus 4. And that’s our answer. But let’s figure out all of the other points. So, B – you go up 2 and to the right 3 so that you go from the new C, you would go, let me make sure. Right. About the origin, right. So the new C, let me just make sure I’m – right. So, the new B would go down 3 down to the left. Okay, that’d be there. So, that’s the new B, right? And so, that looks right. That’s what the triangle is going to look like. This is B prime. This is A prime. This is C prime, right? This is the new triangle when you rotate it 180 degrees.

That was a challenging visualization problem. But the key thing is just to keep visualizing going around 180 degrees. So minus 5, minus 4, that’s choice A for A prime.

Problem 80. Now, you switch to better color. Trapezoid ABCD below is to be translated to trapezoid A prime, B prime, C prime, D prime by the following motion rule. Okay? What will be their coordinates of vertex C prime? So, that’s what we have to worry about. We’re at C and we’re going to translate it to some C prime. What are the coordinates of C? C is – X is equal to 5 and y is equal to 1. So, what’s C prime going to be equal to? C prime, we’ll just use this mapping rule, you add 3 to the x. That’s going to be 8. And you subtract 4 from the y. So, 8 – 1 minus 4 is minus 3. 8 minus 3. And that’s choice D.

And we’re all done. We’re all done with the entire California standards Geometry Class. And frankly, you know, I started off grumbling a little bit because I thought they were getting a little bit too obsessed with the terminology, but in over all, I think this was a pretty good test. And if you understood all the problems on this exam, I think you know your Geometry pretty well.

Anyway, see you in the next video although this is the last in Geometry – in this Geometry series.