In today's class, we had a pre-test. We had at least 20 minutes to finish the work. Then we were off to groups.

Here were the questions:

This question can be proved by the tangent chord theorem.

"The angle between a tangent and a chord is congruent to the inscribed angle subtended by the opposite side of the chord."

As you can see we have the AB as our tangent. We also have angle r and angle d as our angles between the chord and the tangent. Angle a is in the opposite side of the chord of angle r, therefore angle r is congruent to angle a.

Next question:

Since OB and OD are radii, they must be congruent. Therefore, we can form an isosceles triangle through ΔOBD. We then formed a right angle through angle OBC because OB is perpendicular to AC. We can prove this by the tangent-radius theorem. Now that we've formed a right angle through angle OBC, we can now find angle OBD, which is 50° by subtracting 40°(given) from 90°(right angle). We've also said that the triangle was an isosceles triangle so angle OBD must be congruent to angle BDO. Therefore OBD and BDO are 50°. We can now find angle x through the sum of angles in ΔDBO, 180° - (50° + 50°) = 80°.

Next question

This one is pretty straight forward. We can find angle x by doing Thales' theorem.

"If an inscribed angle is subtended by the diameter then the inscribed angle is a right angle."

Now that we know, there is a right triangle, we can now easily find angle x. We can now find x by subtracting 40°(given) from 90°(right angle) which is equal to 50°. Therefore x is equal to 50°.

Next question:It shows that we've got 2 isosceles triangle, ΔRSU and ΔSTU. Lets focus on eliminating y. In ΔRSU, we have y + 2a = 180°. In a cyclic quadrilateral RSUT, we have y + b = 180°. We can isolate y here, so we get y = 180° - b. Now we can substitute the y in therefore we get (180° - b) + 2a = 180°. We now get 2a = b.

We can also say that this cyclic quadrilateral is an isosceles trapezoid. We can prove this by saying RS is congruent to UT therefore we can say that RU and ST are parallel. Since they are isosceles we can say that x = b - a. Since we said b = 2a then we can substitute b into 2a. Now we get x = 2a - a. Then we get x = a.

We can now find x through b + 2a + x = 180. By substituting what we found the equation would turn out to be 2a + 2a + a = 180. This would equal to 5a = 180. Therefore we've got a = 36°. We said that a = x, therefore x is also 36°.

## Wednesday, December 10, 2008

Subscribe to:
Post Comments (Atom)

## 1 comment:

That last question seemed like a doozy, great job on getting the answer :D

Post a Comment