2008 AMC 12A Problems/Problem 24: Difference between revisions
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==Problem== | |||
Triangle <math>ABC</math> has <math>\angle C = 60^{\circ}</math> and <math>BC = 4</math>. Point <math>D</math> is the midpoint of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>? | |||
<math>\textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} - 3} \qquad \textbf{(E)}\ 1</math> | |||
==Solution== | |||
==See Also== | |||
{{AMC12 box|year=2008|ab=A|num-b=22|num-a=24}} | |||
Revision as of 21:43, 22 February 2008
Problem
Triangle 
has 
and 
. Point 
is the midpoint of 
. What is the largest possible value of 
?
Solution
See Also
| 2008 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 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 | |
