In [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) $ABC$, $\tan \angle CAB = 22/7$, and the [altitude](https://artofproblemsolving.com/wiki/index.php/Altitude) from $A$ divides $BC$ into [segments](https://artofproblemsolving.com/wiki/index.php/Segment) of length 3 and 17.  What is the area of triangle $ABC$?
[AIME 1988 Solution 07.png](https://artofproblemsolving.com/wiki/index.php/File:AIME_1988_Solution_07.png)
Call $\angle BAD$ $\alpha$ and $\angle CAD$ $\beta$. So, $\tan \alpha = \frac {17}{h}$ and $\tan \beta = \frac {3}{h}$. Using the tangent addition formula $\tan (\alpha + \beta) = \dfrac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}$, we get $\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 - 51}{h^2}} = \frac {22}{7}$. 
Simplifying, we get $\frac {20h}{h^2 - 51} = \frac {22}{7}$. Cross-multiplying and simplifying, we get $11h^2-70h-561 = 0$. Factoring, we get $(11h+51)(h-11) = 0$, so we take the positive positive solution, which is $h = 11$. Therefore, the answer is $\frac {20 \cdot 11}{2} = 110$, so the answer is $\boxed {110}$.
~Arcticturn