Question 18 of 20 This test 20 point(s) possible This question: 1 point(s) possible Find one solution for the equation. Assume that all angles involved are acute angles. \[ \tan \left(2 B-15^{\circ}\right)=\cot \left(4 B+15^{\circ}\right) \] \[ \mathrm{B}=\square^{\circ} \] (Simplify your answer.)

To solve the equation \[ \tan(2B - 15^\circ) = \cot(4B + 15^\circ), \] we can use the identity that relates tangent and cotangent: \[ \cot(x) = \frac{1}{\tan(x)}. \] Thus, we can rewrite the equation as: \[ \tan(2B - 15^\circ) = \frac{1}{\tan(4B + 15^\circ)}. \] This implies: \[ \tan(2B - 15^\circ) \tan(4B + 15^\circ) = 1. \] Using the identity \(\tan(A) \tan(B) = 1\) leads us to the conclusion that: \[ 2B - 15^\circ + 4B + 15^\circ = 90^\circ + n \cdot 180^\circ, \] for some integer \(n\). Simplifying this gives: \[ 6B = 90^\circ + n \cdot 180^\circ. \] Now, we can solve for \(B\): \[ B = \frac{90^\circ + n \cdot 180^\circ}{6}. \] To find acute angles, we will consider \(n = 0\): \[ B = \frac{90^\circ}{6} = 15^\circ. \] Now, let's check if \(n = 1\) gives us an acute angle: \[ B = \frac{90^\circ + 180^\circ}{6} = \frac{270^\circ}{6} = 45^\circ. \] For \(n = 2\): \[ B = \frac{90^\circ + 360^\circ}{6} = \frac{450^\circ}{6} = 75^\circ. \] For \(n = 3\): \[ B = \frac{90^\circ + 540^\circ}{6} = \frac{630^\circ}{6} = 105^\circ, \] which is not acute. Thus, the acute angle solutions we found are \(15^\circ\), \(45^\circ\), and \(75^\circ\). The simplest solution is: \[ \mathrm{B} = 15^\circ. \] So, the final answer is: \[ \mathrm{B} = 15^\circ. \]