Difference between revisions of "2008 AMC 12A Problems/Problem 24"

Revision as of 22:43, 22 February 2008

Problem

Triangle $ABC$ has $\angle C = 60^{\circ}$ and $BC = 4$ . Point $D$ is the midpoint of $BC$ . What is the largest possible value of $\tan{\angle BAD}$ ?

$\textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} - 3} \qquad \textbf{(E)}\ 1$

Solution

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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

Revision as of 22:43, 22 February 2008 (view source) Xantos C. Guin (talk \| contribs) ← Older edit		Revision as of 22:43, 22 February 2008 (view source) Xantos C. Guin (talk \| contribs) m (typo) Newer edit →
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	==See Also==		==See Also==
−	{{AMC12 box\|year=2008\|ab=A\|num-b=22\|num-a=24}}	+	{{AMC12 box\|year=2008\|ab=A\|num-b=23\|num-a=25}}

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Difference between revisions of "2008 AMC 12A Problems/Problem 24"

Revision as of 22:43, 22 February 2008

Problem

Solution

See Also