Difference between revisions of "1995 AHSME Problems/Problem 18"
(New page: ==Problem== Two rays with common endpoint <math>O</math> forms a <math>30^\circ</math> angle. Point <math>A</math> lies on one ray, point <math>B</math> on the other ray, and <math>AB = 1<...) |
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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> | ||
+ | |||
+ | <math>2\sin{\angle A}=OB</math>. | ||
+ | |||
+ | 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== | ||
+ | {{AHSME box|year=1995|num-b=17|num-a=19}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:59, 5 July 2013
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.