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

(Solution)
(Solution)
Line 11: Line 11:
 
By the [[Law of Cosines]],
 
By the [[Law of Cosines]],
 
<cmath>\begin{align*}
 
<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>
 
\end{align*}</cmath>
  

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 $\frac{\pi/3}{\pi} = \frac 13 \Rightarrow \mathrm{(D)}$.

See also

2003 AMC 12B (ProblemsAnswer KeyResources)
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

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