In a triangle ABC,if 3 sin A + 4 cos B = 6 an | vedclass

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(B) Given,$3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$. Squaring and adding both equations,we get: $(3 \sin A + 4 \cos B)^2 + (4 \sin B + 3

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$\frac{1}{2}$

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(B) Given,$3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$. Squaring and adding both equations,we get: $(3 \sin A + 4 \cos B)^2 + (4 \sin B + 3 \cos A)^2 = 6^2 + 1^2$ $9 \sin^2 A + 16 \cos^2 B + 24 \sin A \cos B + 16 \sin^2 B + 9 \cos^2 A + 24 \sin B \cos A = 37$ $9(\sin^2 A + \cos^2 A) + 16(\sin^2 B + \cos^2 B) + 24(\sin A \cos B + \cos A \sin B) = 37$ Using the identities $\sin^2 	heta + \cos^2 	heta = 1$ and $\sin(A + B) = \sin A \cos B + \cos A \sin B$: $9(1) + 16(1) + 24 \sin(A + B) = 37$ $25 + 24 \sin(A + B) = 37$ $24 \sin(A + B) = 12$ $\sin(A + B) = \frac{12}{24} = \frac{1}{2}$

In a triangle $ABC$,if $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$,then $\sin (A + B)$ is equal to

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