Math Class - 3/Apr./2009 - Summary

So, class begun with us correcting the homework, and as soon as we were done with that we went on to learning the new subject from the next lesson. Here's what we learned:
1-Secants
2-Theorem 9-11
3-Theorem 9-12
3-Theorem 9-13

So firstly, what is a secant? We've already studied radii and diameters which are segments inside the circle which intersect it at one and two points respectively. We've also studied tangents already, which is a segment or line that intersect the circle at exactly ONE point on the circle. So, by common sense, a secant is obviously a kind of fusion of these two concepts, it's a LINE that comes from outside the circle, enters it and leaves it, intersecting the circle at exactly TWO points. In other words, it's nothing more than the continuation of a chord. The image to the right shows what a secant is. In the image, as you can see, line a or BC is a secant to cirlcle A.

After this, we learned THEOREM 9-11. Theorem 9-11 talks about a secant and a tangent meeting on the circle, at exactly the point of tangency. For the nuckleheads who have forgotten what a point of tangency is, it's the point on the circle where the tangent meets the circle, it will ALWAYS be ON the circle, for the obvious reason that a tangent can't be in the circle. The image at the right illustrates the situation proposed by theorem 9-11. Notice how line a or DC is a tangent and how line b or EC is a secant. Also, notice how both intersect at exactly the point of tangency and form 4 angles, with 2 measures, for from these 4 angles, 2 are vertical, tehrefore congruent (theorem from chapter 2 I think, I'm without the book here so I can't say for sure if it's theorem 2-7). So, the theorem says that having this situation where a tangent and a secant intersect at the point of tangency forming 4 angles, then the measure of each angle (angles towards the cirlce) is half of it's bisected corresponding arc. In the image to the right for example, the measure of angle ECF is axactly half the measurment of arc EC, the same works to angle ECD, which is half of arc EBC. This works for a simple reason: as we know, the measurments of all arcs in a circle (in this case 2) add up to 360 degrees, and because angle ECF and angle ECD form a linear pair, they add up to 180 degrees (also chapter 2, by theorem 2-2 or 2-3). In easy words, in a if-then statement the thorem goes like this:
If- A secant and a tangent meet at the point of tangency.
Then- Each angle formed is half of it's bisected corresponding arc.

Moving on, we went to learn THEOREM 9-12. Theorem 9-12 talks about two secants meeting INSIDE the circle at any point. For this theorem, the secants also form 4 angles, but only two measures. The image to the right illustrates the situation proposed by this theorem. In the image, CB and DE are secants, intersecting inside the circle, at point F (the black point). To find the measure of any of these four angles, add up the arcs formed and divide it by two, not the addition of all arcs, but the addition of the corresponding arcs to the angle you want to find and its vertical angle. I know it sounds very hard, but its pretty easy to figure it out, follow with me. When the secants intercept, four angles are formed, where two of them are vertical, in the figure to the right for example, angles BFE, BFD, DFC, and CFE are formed, where EFC and BFD are vertical, and so are EFB and DFC. If you want to find the measurment of BFD for example, don't forget to take in consideration its vertical angle, which is EFC. Both of these angles interect arcs, take the measurments of these and add them, then divide the sum by 2. in the example angles BFD and EFC= (m arc EC+m arc BD)/2. In a simple if-then statement using the figure, it goes like this:
If- 2 secants meet inside a circle.
Then-
THEOREM 9-13. If theorem 9-11 talked about a point on the circle, theorem 9-12 talked about a point inside the circle, it was obvious that theorem 9-13 would talk about a point outside the circle. For theorem 9-13 there were 3 possible situations being described, the intersection of 2 secants outside the circle, the intersection of 1 tangent and 1 secant outside the circle, and the intersection of two tangents, also outside the circle. For each possibility, I will use a figure to illustrate.

TWO SECANTS:

To the right, you can see in the illustration where DE and BC are secants to circle A. Both intersect outside the circle at F. To find the measurment of angle EFC, the same formula used in theorem 9-12 is used, but instead of adding the measurment of the arcs, you subtract them. The positive difference is divided by two, giving you the exact measurment of the arc. Just don't forget to always subtract the smaller arc from the bigger arc, and not the bigger from the smaller so that you always get a positive number. In the example, angle EFC=(arc BD-arc EC)/2. If you want to find any of the other 3 angles, you can, if its the vertical angle, it has the same measurment, and if it's a linear pair just subtract the measurment from 180.

1 SECANT AND 1 TANGENT:

This theorem also works when there is one secant and one tangent intersecting outside the circle, as shown in the illustration to the right. As you can notice, BC is a secant to circle A, and DF is a tangent to circle A. Both intersect outside the circle at F. Notice how both lines form two arcs, divided by the point of tangency. The same formula cited above in the 2 secants example works, (the larger arc-the smaller arc) divided by two. In this illustration, to find the blue angle (DFC), subtract the yellow arc from the red arc and divide by two.

2 TANGENTS


Finally, the last situation proposed by this theorem is of two tangents intersecting outside the circle (obviously outside the circle because tangents cannot go inside a circle). The illustration to the right shows this proposed situation. As you can notice (actually you can't notice because I forgot to make the font larger, sorry about that), lines CD and EF are tangents to circle A, intersecting at G. Notice how the points of tangency divide the circle into exactly two arcs, a major and a minor arc. Using this information, we can take the same formula used before in the previous situations to figure ou the measure of angle CGE, in the illustration, its the red arc minus the yellow arc, all divided by two. Another important information is to not forget to include a third point in the major arc so that you can diferentiate it from the minor arc, in the illustration, point B works as the diferentiator so that instead of the formula being arc CE-arc CE, its arc CBE-arc CE. For this theorem, Ms. Saatkamp gave us an example to try to figure out. If in the illustration above, the blue angle measured 56 degrees and the yellow arc measured 56 degrees, what would be the measurment of the red arc? Pamella figured it out almost instantly, because the sum of both arcs is 360 degrees, the formula would go like this: 56=(360-x)-x (assume all numbers are in degrees). If you solved for x using simple 6 grade algebra, your answer should be 152 degrees.

Well, after this we just packed up to go home because it was FRIDAAAYYYY and because the other students are having fun playing like, 4 times a day in the tournament, we don't have any homework hurah! Monday's class will just be doing the homework in class I think, so if it's that we probably don't have homework on monday to (!!).

That's all, see ya'll on monday.