If 5 cos x + 12 cos y = 13,then the maximum v | vedclass

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(B) Given $5 \cos x + 12 \cos y = 13$. Squaring both sides,we get $(5 \cos x + 12 \cos y)^2 = 169$. Let $S = 5 \sin x + 12 \sin y$. Consider $S^2 + 13

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$\sqrt{120}$

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(B) Given $5 \cos x + 12 \cos y = 13$. Squaring both sides,we get $(5 \cos x + 12 \cos y)^2 = 169$. Let $S = 5 \sin x + 12 \sin y$. Consider $S^2 + 13^2 = (5 \sin x + 12 \sin y)^2 + (5 \cos x + 12 \cos y)^2$. $= 25(\sin^2 x + \cos^2 x) + 144(\sin^2 y + \cos^2 y) + 120(\sin x \sin y + \cos x \cos y)$. $= 25 + 144 + 120 \cos(x - y) = 169 + 120 \cos(x - y)$. Thus,$S^2 = 169 + 120 \cos(x - y) - 169 = 120 \cos(x - y)$. Since the maximum value of $\cos(x - y)$ is $1$,the maximum value of $S^2$ is $120$. Therefore,the maximum value of $S$ is $\sqrt{120}$.

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

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