Difference between revisions of "2011 AMC 12B Problems/Problem 6"
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==Solution 2== | ==Solution 2== | ||
| − | Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is <math>\frac{3x-2x}2 = \frac {x} 2</math> by the arc length formula. Also note that <math>3x + 2x = | + | Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is <math>\frac{3x-2x}2 = \frac {x} 2</math> by the arc length formula. Also note that <math>3x + 2x = 180</math>, which simplifies to <math>x= 72.</math> Hence the angle formed by the tangents is equal to <math>\boxed{36 \textbf{(C)}}</math>. $ |
==See also== | ==See also== | ||
{{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}} | {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:19, 3 January 2019
Contents
Problem
Two tangents to a circle are drawn from a point 
. The points of contact 
and 
divide the circle into arcs with lengths in the ratio 
. What is the degree measure of 
?
Solution
In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).
In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d.
Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.
Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:
1/2 (216°-144°) = 1/2 (72°) 
Solution 2
Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is 
by the arc length formula. Also note that 
, which simplifies to 
Hence the angle formed by the tangents is equal to 
. $
See also
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
