Question 2 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. \[ \sec \left(3 \beta+16^{\circ}\right)=\csc \left(\beta-2^{\circ}\right) \] \[ \beta=\square^{\circ} \] (Simplify your answer.)

To solve the equation \(\sec \left(3 \beta + 16^{\circ}\right) = \csc \left(\beta - 2^{\circ}\right)\), we first rewrite secant and cosecant in terms of sine and cosine: \[ \frac{1}{\cos(3\beta + 16^{\circ})} = \frac{1}{\sin(\beta - 2^{\circ})} \] This leads to the equation: \[ \sin(\beta - 2^{\circ}) = \cos(3\beta + 16^{\circ}) \] Using the co-function identity \(\sin(x) = \cos(90^{\circ} - x)\), we rewrite the equation as: \[ \cos(90^{\circ} - (\beta - 2^{\circ})) = \cos(3\beta + 16^{\circ}) \] This means we can set up two equations based on that: 1. \(90^{\circ} - (\beta - 2^{\circ}) = 3\beta + 16^{\circ}\) 2. \(90^{\circ} - (\beta - 2^{\circ}) = -(3\beta + 16^{\circ})\) ### Solving the first equation: From equation 1: \[ 90^{\circ} - \beta + 2^{\circ} = 3\beta + 16^{\circ} \] Combining like terms: \[ 92^{\circ} - \beta = 3\beta + 16^{\circ} \] \[ 92^{\circ} - 16^{\circ} = 4\beta \] \[ 76^{\circ} = 4\beta \] \[ \beta = 19^{\circ} \] ### Solving the second equation: From equation 2: \[ 90^{\circ} - \beta + 2^{\circ} = -3\beta - 16^{\circ} \] \[ 92^{\circ} - \beta = -3\beta - 16^{\circ} \] Rearranging gives: \[ 92^{\circ} + 16^{\circ} = -3\beta + \beta \] \[ 108^{\circ} = -2\beta \] \[ \beta = -54^{\circ} \] Since we require acute angles, we disregard \(-54^{\circ}\). Thus, the acute solution is: \[ \beta = 19^{\circ} \] Therefore, the answer is: \[ \beta = 19^{\circ} \]