# 2011 AMC 12B Problems/Problem 6

## 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

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