Question

If $5 \cos x+12 \cos y=13$, then the maximum value of $5 \sin x+12 \sin y$ is :

• A. $12$
• B. $\sqrt{120}$
• C. $\sqrt{20}$
• D. $13$

Answer 1

Answer: (B)

$\because 5 \cos x+12 \cos y=13$
$\Rightarrow (5 \cos x+12 \cos y)^{2}=169$
$\therefore (5 \cos x+12 \cos y)^{2}+(5 \sin x+12 \sin y)^{2}$
$\Rightarrow (13)^{2}+(5 \sin x+12 \sin y)^{2}$
$=25+144+120(\sin x \sin y+\cos x \cos y)$
$\Rightarrow (5 \sin x+12 \sin y)^{2}=120 \cos (x-y)$
$\because -1 \leq \cos (x-y) \leq 1$
$\therefore $ Maximum value of $5 \sin x+12 \sin y$
$=\sqrt{120}$