Difference between revisions of "1999 AHSME Problems/Problem 27"
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Revision as of 14:35, 5 July 2013
Problem
In triangle ,
and
. Then
in degrees is
Solution
Square the given equations and add (simplifying with the Pythagorean identity ):
Thus . This is the sine addition identity, so
. Thus either
.
If , then
, and
. The first equation implies
, which is a contradiction; thus
.
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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 |
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