Difference between revisions of "1988 AIME Problems/Problem 7"
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| − | Call <math>\angle BAD</math> <math>\alpha</math> and <math>\angle CAD</math> <math>\beta</math>. So, <math>\tan \alpha = \frac {17}{h}</math> and <math>\tan \beta = \frac {3}{h}</math>. Using the tangent addition formula <math>\tan (\alpha + \beta) = \frac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \ beta}</math>, we get <math>\ | + | Call <math>\angle BAD</math> <math>\alpha</math> and <math>\angle CAD</math> <math>\beta</math>. So, <math>\tan \alpha = \frac {17}{h}</math> and <math>\tan \beta = \frac {3}{h}</math>. Using the tangent addition formula <math>\tan (\alpha + \beta) = \frac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \ beta}</math>, we get <math>\dfrac {\frac {20}{h}}{\frac {x^2 - 51}{x^2}}</math>. |
== See also == | == See also == | ||
Revision as of 23:20, 20 November 2023
Problem
In triangle 
, 
, and the altitude from 
divides 
into segments of length 3 and 17. What is the area of triangle ?
Solution
Call 
and 
. So, 
and 
. Using the tangent addition formula 
, we get 
.
See also
| 1988 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
