Answer: \(\triangle CPA \sim \triangle CQB\) by the AA test, and AC = 10.5
Proof: In \(\triangle CPA\) and \(\triangle CQB\):
- \(\angle CPA \cong \angle CQB\) (each \(90^\circ\), since AP ⊥ BC and BQ ⊥ AC)
- \(\angle ACP \cong \angle BCQ\) (common angle C)
\(\therefore \triangle CPA \sim \triangle CQB\) (AA test).
Finding AC: corresponding sides of similar triangles are in proportion:
\[\frac{AP}{BQ} = \frac{AC}{BC}\] \[\frac{7}{8} = \frac{AC}{12}\] \[AC = \frac{7 \times 12}{8} = 10.5\]