A) 12
B) \[\sqrt{120}\]
C) \[\sqrt{20}\]
D) 13
Correct Answer: B
Solution :
Since,\[5\cos x+12\cos y=13\] \[\Rightarrow \]\[{{(5\cos x+12\cos y)}^{2}}=169\] Now, \[{{(5\cos x+12\cos y)}^{2}}+{{(5\sin x+12\sin y)}^{2}}\] \[={{(13)}^{2}}+{{(5\sin x+12\sin y)}^{2}}\] \[\Rightarrow \]\[25+44+120(\sin x\sin y+\cos x\cos y)\] \[=169+{{(5\sin x+12\sin y)}^{2}}\] \[\Rightarrow \] \[{{(5\sin x+12\sin y)}^{2}}=120\cos (x-y)\] \[\because \] \[-1\le \cos (x-y)\le 1\] \[\Rightarrow \] \[-120\le 120\cos (x-y)\le 120\] \[\therefore \] Maximum value of \[5\sin x+12\sin y=\sqrt{120}\]
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