NIMCET 2017 — Mathematics PYQ

Calculation:

Given: a sin A + b cos A = c

Divide both sides by $\frac{1}{\sqrt{a^2+b^2}}$, we get

$\Rightarrow \frac{a}{\sqrt{a^2+b^2}}\sin A+\frac{b}{\sqrt{a^2+b^2}}\cos A=\frac{c}{\sqrt{a^2+b^2}}$

Let $\sin \alpha =\frac{a}{\sqrt{a^2+b^2}}$ and $\cos \alpha =\frac{b}{\sqrt{a^2+b^2}}$

$\Rightarrow \sin A\sin \alpha +\cos A\cos \alpha =\frac{c}{\sqrt{a^2+b^2}}$

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

$\Rightarrow A-\alpha ={\cos }^{-1}\frac{c}{\sqrt{a^2+b^2}}$

$\Rightarrow A={\cos }^{-1}\frac{c}{\sqrt{a^2+b^2}}+\alpha$

Now, $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{a}{b}$

$\therefore \alpha ={\tan }^{-1}\frac{a}{b}$

So, $A={\cos }^{-1}\frac{c}{\sqrt{a^2+b^2}}+{\tan }^{-1}\frac{a}{b}$

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