Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Let

$R = \sqrt{a^2 + b^2}$

Choose $\theta$ such that

$\cos \theta = \frac{b}{R}, \quad \sin \theta = \frac{a}{R}$

Then

$a\sin A + b\cos A = R(\sin A \sin \theta + \cos A \cos \theta) = R\cos(A - \theta)$

Given that

$a\sin A + b\cos A = c,$

we get

$R\cos(A - \theta) = c \quad \Rightarrow \quad \cos(A - \theta) = \frac{c}{\sqrt{a^2 + b^2}}$

Hence,

$A - \theta = \pm \cos^{-1}\!\left(\frac{c}{\sqrt{a^2 + b^2}}\right) + 2\pi n, \quad n \in \mathbb{Z}$

and since $\tan \theta = \dfrac{a}{b}$, we have $\theta = \tan^{-1}\!\left(\dfrac{a}{b}\right)$

Therefore,

$\boxed{A = \tan^{-1}\!\left(\frac{a}{b}\right) \pm \cos^{-1}\!\left(\frac{c}{\sqrt{a^2 + b^2}}\right)}$

Solution

Solution

Solution