In triangle $ABC,$ $BC = 32,$ $\tan B = \frac{3}{2},$ and $\tan C = \frac{1}{2}.$  Find the area of the triangle.
To find the area of the triangle, I need to know its base and height. I could use any side as the base, but I'll choose $BC$ since I already know its length. To find the height, I need to draw an altitude from $A$ to $BC.$ Let $D$ be the foot of the altitude, so $AD$ is perpendicular to $BC.$ Then, by the definition of tangent, I have $\tan B = \frac{AD}{BD}$ and $\tan C = \frac{AD}{CD}.$ Multiplying both equations by $BD$ and $CD$ respectively, I get $AD = BD \tan B$ and $AD = CD \tan C.$ Since $AD$ is the same in both equations, I can equate them and get $BD \tan B = CD \tan C.$ I can also use the fact that $BD + CD = BC = 32$ to eliminate one of the variables. For example, if I solve for $BD$ in terms of $CD,$ I get $BD = 32 - CD.$ Substituting this into the equation involving tangents, I get $(32 - CD) \tan B = CD \tan C.$ Expanding and rearranging, I get $CD (\tan B + \tan C) = 32 \tan B.$ Dividing both sides by $\tan B + \tan C,$ I get $CD = \frac{32 \tan B}{\tan B + \tan C}.$ I can plug in the values of $\tan B = \frac{3}{2}$ and $\tan C = \frac{1}{2}$ and simplify. I get $CD = \frac{32 \cdot \frac{3}{2}}{\frac{3}{2} + \frac{1}{2}} = \frac{96}{4} = 24.$ Now that I have $CD,$ I can find $BD$ by subtracting it from $32.$ I get $BD = 32 - 24 = 8.$ To find the height $AD,$ I can use either equation involving tangents. For example, I can use $AD = CD \tan C.$ I get $AD = 24 \cdot \frac{1}{2} = 12.$ Finally, I have the base and the height of the triangle, so I can find the area by using the formula $A = \frac{1}{2}bh.$ I get $A = \frac{1}{2} \cdot 32 \cdot 12 = 192.$

