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

Revision as of 01:39, 26 January 2021

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°) $=\boxed{36 \textbf{(C)}}.$

Solution 2

Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is $\frac{3x-2x}2 = \frac {x} 2$ by the arc length formula. Also note that $3x + 2x = 360$ , which simplifies to $x= 72.$ Hence the angle formed by the tangents is equal to $\boxed{36 \textbf{(C)}}$ .

Solution 3

Let the center of the circle be $O$ . The radii extending from $O$ to the points of tangency $B$ and $C$ are both perpendicular to lines $AB$ and $AC$ . Thus $\angle ABO = \angle ACO = 90^{\circ}$ , and quadrilateral $ABOC$ is cyclic, since the opposite angles add to $180^{\circ}$ .

Since the ratio of the arc lengths is $2 : 3$ , the shorter arc (the arc cut off by $\angle BOC$ ) is $144^{\circ}$ .

$\angle BOC$ and $\angle BAC$ must add up to $180^{\circ}$ since quadrilateral $ABOC$ is cyclic, so $\angle BAC = 180 - 144 = \boxed{\textbf{(C) } 36}$ .

-Solution by Joeya

@@ Line 26: / Line 26: @@
 <math>\angle BOC</math> and <math>\angle BAC</math> must add up to <math>180^{\circ}</math> since quadrilateral <math>ABOC</math> is cyclic, so <math>\angle BAC = 180 - 144 = \boxed{\textbf{(C) } 36}</math>.
+-Solution by Joeya
 ==See also==
 {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}}
 {{MAA Notice}}

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

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