Difference between revisions of "2011 AMC 12B Problems/Problem 6"

(Solution)
(Solution)
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Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:
 
Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:
  
1/2 (216°-144°) = 1/2 (72°) <math> =\boxed{36 \textbf{(C)}} </math>
+
1/2 (216°-144°) = 1/2 (72°) <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 16:10, 19 December 2018

Problem

Two tangents to a circle are drawn from a point $A$. The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$. What is the degree measure of $\angle{BAC}$?

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$

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°) $=\boxed{36 \textbf{(C)}}.$

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
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

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