### Instruction: In triangle $ABC$, $\tan \angle CAB = \frac{22}{7}$, and the altitude from $A$ divides $\overline{BC}$ into segments of length $3$ and $17$. What is the area of triangle $ABC$?### Response: I need to find the area of triangle $ABC$, which is half the product of the base and the height.
The base is $\overline{BC}$, which has length $3 + 17 = 20$.
The height is the length of the altitude from $A$.
To find the altitude, I can use trigonometry and the fact that $\tan \angle CAB = \frac{22}{7}$.
I know that $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$, where $\theta$ is an acute angle in a right triangle.
In this case, $\theta = \angle CAB$, the opposite side is the altitude, and the adjacent side is one of the segments of $\overline{BC}$.
I can choose either segment, but I will pick the one with length $17$, since it is closer to the tangent value.
So, $\tan \angle CAB = \frac{\text{altitude}}{17}$, and I can solve for the altitude by multiplying both sides by $17$.
I get $\text{altitude} = 17 \tan \angle CAB = 17 \cdot \frac{22}{7} = \frac{374}{7}$.
Now I have the base and the height of the triangle, so I can find the area by multiplying them and dividing by $2$.
The area is $\frac{1}{2} \cdot 20 \cdot \frac{374}{7} = \frac{3740}{7}$.