Tuesday, December 9, 2008

Today's Class

For today's class, we had a workshop class instead of a pre-test because we decided we needed one more day to prepare for the pre-test and test.
We were to work on the following questions and were to solve for Angle 1 and Angle 2. Here are the questions. I'll post the answers to each of them separately. I'll also provide an explanation for each.

For circle one:

To find Angle 2 you just need to know if two angles are opposite of each other in a polygon then the sum of those two angles equal to 180° by the cyclic quadrilateral theorem. Since 100° is given, and it's opposite angle is 2 you just need to subtract 100° from 180° to get angle 2, which is 80°

We had a discussion for finding Angle 1. We were to find the reason why lines AD and BC are parallel. We used the Parallel Chords Theorem in a way to explain why AD and BC are congruent. Since AB and DC are congruent, we can say that the Arc's subtending lines AB and DC are congruent. That means if the Arc's are the same, the lines AD and BC must be parallel by the Parallel chords Theorem. Now we can solve for Angle 1. Since lines AB and DC are congruent and lines AD and BC are parallel, we get an Isosceles Trapezoid. Meaning two base angles are the same, therefore two sides are also the same. Since it's given lines AB and DC are congruent, we can now say Angle 1 is 80°.

For circle two:
To find Angle 1 we first need to find angle BCD. Since angle BCE is given, and DE is a straight line, we need to find the supplementary angle to angle BCE. To find angle BCD we subtract 100° from 180° to give us angle BCD, which is 80°. We can now find Angle 1 by using the Cyclic quadrilateral theorem. Angle 1 is opposite of angle BCD, therefore the sum of those two angles add up to 180°. Angle 1 one is 100° by subtracting 80° from 100°.

For circle three:

The first step to do is to find Angle CDA. We find this angle by subtracting the sum of the known angles from 180°. Angle ACD is a right angle and angle CDA is given ( 25° ), Angle CDA is 65°.

Since ABCD is a Cyclic quadrilateral, we can use the Cyclic quadrilateral theorem to find Angle ABC. Opposite to angle ABC is angle CDA, which is 65°. ABC and CDA must me supplementary, therefore Angle ABC is 115°.

It's given that lines AB and BC are congruent, therefore an Isosceles triangle is formed, triangle ABC. That also means Angles BAC and BCA are congruent. So all we have to do now is to subtract 115° from 180° and divide by 2 to get those 2 angles. Leaving us with 65° divided by 2 which equals to 32.5°. 1 = 32.5°

We also had a second question. We didn't have much time to solve it, but I'll post the question for those who need it.

I'll also post my solution to the question:

Note: My solution may contain errors. I've provided my solution to give an idea / jump start to those who may be stuck. Don't take it as the 100% correct answer.

That sum's up the class for today. Remember we have a pre-test tomorrow and the test on Thursday. Study hard, and good luck!

Next Scribe: jdizon

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