Difference between revisions of "1992 AHSME Problems/Problem 25"
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== Problem == | == Problem == | ||
| − | In <math>\triangle{ABC}</math>, <math>\angle | + | In <math>\triangle{ABC}</math>, <math>\angle ABC=120^\circ,AB=3</math> and <math>BC=4</math>. If perpendiculars constructed to <math>\overline{AB}</math> at <math>A</math> and to <math>\overline{BC}</math> at <math>C</math> meet at <math>D</math>, then <math>CD=</math> |
<math>\text{(A) } 3\quad | <math>\text{(A) } 3\quad | ||
Revision as of 19:27, 9 April 2015
Problem
In 
, 
and 
. If perpendiculars constructed to 
at 
and to 
at 
meet at 
, then 
Solution
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 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. 
