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*For number six, assume that the line M is not a tangent, meaning it intersects the circle in more than one point. Call the points of intersection A and X. The theorem assumes that angle OAX is a right angle. But consider lengths OX and OA: they are both radii, and so they are equal. Viewing OAX as a triangle, what impossible circumstance does this last statement imply? | *For number six, assume that the line M is not a tangent, meaning it intersects the circle in more than one point. Call the points of intersection A and X. The theorem assumes that angle OAX is a right angle. But consider lengths OX and OA: they are both radii, and so they are equal. Viewing OAX as a triangle, what impossible circumstance does this last statement imply? | ||
| + | For 7 use THM 38 and the fact that there are 180 degrees in a line! | ||
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Revision as of 13:07, 18 November 2009
HW 11
Does anyone have any hints for 6, 8, or 10?
for 10, draw out all of the little triangles separately. Start marking which sides equal which sides. you eventually get to triangle rpq equal to triangle spq so congruent angles.
I don't know those, but what about 3,7, and 9?
for 3, do law of sines 4 times (3 little triangles and big triangle) and simplyfy down.
for 9, use thm 17 3 times.
I'm still stumped on 6 and 7.
- For number six, assume that the line M is not a tangent, meaning it intersects the circle in more than one point. Call the points of intersection A and X. The theorem assumes that angle OAX is a right angle. But consider lengths OX and OA: they are both radii, and so they are equal. Viewing OAX as a triangle, what impossible circumstance does this last statement imply?
For 7 use THM 38 and the fact that there are 180 degrees in a line!
