Step-by-step Solution:

\[ \text{Given: } a \sin A + b \cos A = c \] Step 1: Normalize the Equation Divide both sides by \( \sqrt{a^2 + b^2} \): \[ \frac{a}{\sqrt{a^2 + b^2}} \sin A + \frac{b}{\sqrt{a^2 + b^2}} \cos A = \frac{c}{\sqrt{a^2 + b^2}} \] Define: \[ \sin \alpha = \frac{a}{\sqrt{a^2 + b^2}}, \quad \cos \alpha = \frac{b}{\sqrt{a^2 + b^2}} \] Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \), this substitution is valid. Now, rewriting the equation: \[ \sin A \sin \alpha + \cos A \cos \alpha = \frac{c}{\sqrt{a^2 + b^2}} \] Using the cosine addition identity: \[ \cos(A - \alpha) = \frac{c}{\sqrt{a^2 + b^2}} \] Step 2: Solve for \( A \) \[ A - \alpha = \cos^{-1} \left( \frac{c}{\sqrt{a^2 + b^2}} \right) \] \[ A = \cos^{-1} \left( \frac{c}{\sqrt{a^2 + b^2}} \right) + \alpha \] Since \( \alpha = \tan^{-1} \left(\frac{a}{b}\right) \), we substitute: \[ A = \cos^{-1} \left( \frac{c}{\sqrt{a^2 + b^2}} \right) + \tan^{-1} \left(\frac{a}{b}\right) \] Final Answer: \[ A = \tan^{-1} \left(\frac{a}{b}\right) + \cos^{-1} \left(\frac{c}{\sqrt{a^2 + b^2}}\right) \]