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: $(5 \cos x + 12 \cos y)^2 = 169$. Let $S = 5 \sin x + 12 \sin y$. Consider $A = (5 \cos x

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$\sqrt{120}$

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

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

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