Difference between revisions of "1995 AHSME Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | Triangle <math>OAB</math> has the property that <math>\angle | + | Triangle <math>OAB</math> has the property that <math>\angle O=30^{\circ}</math> and <math>AB=1</math>. |
| + | |||
| + | From the [[Law of Sines]], <math>\frac{\sin{\angle A}}{OB}=\frac{\sin{\angle O}}{AB}</math>. | ||
| + | |||
| + | Since <math>\sin 30^\circ = \frac{1}{2}</math>, we have: | ||
| + | |||
| + | <math>\frac{\sin{\angle A}}{OB}=\frac{\frac{1}{2}}{1}</math> | ||
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| + | <math>2\sin{\angle A}=OB</math>. | ||
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| + | The maximum of <math>\sin{\angle A}</math> is <math>1</math> when <math>\angle A = 90^\circ</math>, so the maximum of <math>OB</math> is <math>2\Rightarrow \mathrm{(D)}</math>. | ||
==See also== | ==See also== | ||
Revision as of 18:20, 18 August 2011
Problem
Two rays with common endpoint 
forms a 
angle. Point 
lies on one ray, point 
on the other ray, and 
. The maximum possible length of 
is
Solution
Triangle 
has the property that 
and 
.
From the Law of Sines, 
.
Since 
, we have:
.
The maximum of 
is 
when 
, so the maximum of is 
.
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
