Difference between revisions of "2003 AMC 12B Problems/Problem 21"

Revision as of 22:13, 18 August 2016

Problem

An object moves $8$ cm in a straight line from $A$ to $B$ , turns at an angle $\alpha$ , measured in radians and chosen at random from the interval $(0,\pi)$ , and moves $5$ cm in a straight line to $C$ . What is the probability that $AC < 7$ ?

$\mathrm{(A)}\ \frac{1}{6} \qquad\mathrm{(B)}\ \frac{1}{5} \qquad\mathrm{(C)}\ \frac{1}{4} \qquad\mathrm{(D)}\ \frac{1}{3} \qquad\mathrm{(E)}\ \frac{1}{2}$

Solution

By the Law of Cosines, $\begin{align*} AB^2 + BC^2 - 2 AB \cdot BC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\ \cos \alpha &< \frac 12\\ \end{align*}$

It follows that $0 < \alpha < \frac {\pi}3$ , and the probability is $\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}$ .

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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.

@@ Line 11: / Line 11: @@
 By the [[Law of Cosines]],
 <cmath>\begin{align*}
-AB^2 + BC^2 - 2 AB \cdot BC \cos \beta = 89 - 80 \cos \beta = AC^2 &< 49\\
+AB^2 + BC^2 - 2 AB \cdot BC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\
-\cos \beta &< \frac 12\\
+\cos \alpha &< \frac 12\\
 \end{align*}</cmath>

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Difference between revisions of "2003 AMC 12B Problems/Problem 21"

Revision as of 22:13, 18 August 2016

Problem

Solution

See also