Explanation

(c) Here, $y_{1} = 5(\sin{2\pi t} + \sqrt{3}\cos{2\pi t})$
$y_{2} = 5\sin\left( 2\pi t + \frac{\pi}{4} \right)$
$y_{1} = 5\sin{2\pi t + 5\sqrt{3}}\cos{2\pi t}$
As of the form of $y_{1} = \alpha\sin{2\pi t + \beta\cos{2\pi t}}$
Let $\alpha = r\cos\theta = 5,\ \beta = r\sin\theta = 5\sqrt{3}$
$\therefore y_{1} = r\cos\theta\sin{2\pi t + r\sin{\theta\cos{2\pi t}}}$
$= r\sin(2\pi t + \theta)$
Also, $\alpha^{2} + \beta^{2} = r^{2}\cos^{2}{\theta + r^{2}\sin^{2}\theta = r^{2}}$
$\Rightarrow r = \sqrt{\alpha^{2} + \beta^{2}} = \sqrt{(5)^{2} + \left( 5\sqrt{3} \right)^{2}} = 5\sqrt{1^{2} + \left( \sqrt{3} \right)^{2}} = 10$
$\therefore y_{1} = 10\sin{(2\pi t + \theta)}$
$\therefore\ \frac{A_{1}}{A_{2}} = \frac{10}{5} = \frac{2}{1}$