Find one solution for the equation. Assume that all angles involved are acute angles. \( \tan \left(3 \mathrm{~B}-20^{\circ}\right)=\cot \left(2 \mathrm{~B}+15^{\circ}\right) \) \( \mathrm{B}=\square^{\circ} \) (Simplify your answer.)

To solve the equation \(\tan(3B - 20^\circ) = \cot(2B + 15^\circ)\), we can rewrite the cotangent in terms of tangent: \[ \tan(3B - 20^\circ) = \frac{1}{\tan(2B + 15^\circ)} \] Cross-multiplying gives us: \[ \tan(3B - 20^\circ) \tan(2B + 15^\circ) = 1 \] This implies: \[ 3B - 20^\circ + 2B + 15^\circ = 90^\circ + n \cdot 180^\circ \quad \text{(where \(n\) is any integer)} \] Simplifying the left side: \[ 5B - 5^\circ = 90^\circ + n \cdot 180^\circ \] From this, we can rearrange to find \(B\): \[ 5B = 95^\circ + n \cdot 180^\circ \] \[ B = 19^\circ + \frac{n \cdot 180^\circ}{5} \] Since acute angles are between \(0^\circ\) and \(90^\circ\), let's test \(n = 0\): \[ B = 19^\circ \] For \(n = 1\): \[ B = 19^\circ + \frac{180^\circ}{5} = 19^\circ + 36^\circ = 55^\circ \] For \(n = 2\): \[ B = 19^\circ + \frac{360^\circ}{5} = 19^\circ + 72^\circ = 91^\circ \quad \text{(not acute)} \] Thus, the solutions for acute angles \(B\) are: \[ \boxed{19^\circ} \quad \text{and} \quad \boxed{55^\circ} \]