1. **State the problem:** We need to solve the equation $\tan(90^\circ - \theta) = \frac{5}{3}$ for $\theta$. 2. **Recall the formula and identity:** The tangent of the complement angle is related to the cotangent: $$\tan(90^\circ - \theta) = \cot(\theta)$$ Since $\cot(\theta) = \frac{1}{\tan(\theta)}$, the equation becomes: $$\cot(\theta) = \frac{5}{3} \implies \frac{1}{\tan(\theta)} = \frac{5}{3}$$ 3. **Solve for $\tan(\theta)$:** $$\tan(\theta) = \frac{3}{5}$$ 4. **Find $\theta$:** We take the arctangent (inverse tangent) of both sides: $$\theta = \tan^{-1}\left(\frac{3}{5}\right)$$ 5. **Interpretation:** The angle $\theta$ whose tangent is $\frac{3}{5}$ is approximately: $$\theta \approx 30.96^\circ$$ 6. **General solution:** Since tangent is periodic with period $180^\circ$, the general solution is: $$\theta = 30.96^\circ + k \times 180^\circ, \quad k \in \mathbb{Z}$$ **Final answer:** $$\theta \approx 30.96^\circ + 180^\circ k, \quad k \in \mathbb{Z}$$