In triangle ABC,we are given that 3 sin A + 4 | vedclass

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(A) Given equations are:$3 \sin A + 4 \cos B = 6$  $(i)$$4 \sin B + 3 \cos A = 1$  $(ii)$Squaring and adding equations $(i)$ and $(ii)$:$(3 \sin A + 4

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$30$

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(A) Given equations are:$3 \sin A + 4 \cos B = 6$  $(i)$$4 \sin B + 3 \cos A = 1$  $(ii)$Squaring and adding equations $(i)$ and $(ii)$:$(3 \sin A + 4 \cos B)^2 + (4 \sin B + 3 \cos A)^2 = 6^2 + 1^2$$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$$9(\sin^2 A + \cos^2 A) + 16(\sin^2 B + \cos^2 B) + 24(\sin A \cos B + \cos A \sin B) = 37$$9(1) + 16(1) + 24 \sin(A + B) = 37$$25 + 24 \sin(A + B) = 37$$24 \sin(A + B) = 12$$\sin(A + B) = \frac{1}{2}$Since $A + B + C = 180^{\circ}$,we have $A + B = 180^{\circ} - C$.$\sin(180^{\circ} - C) = \frac{1}{2}$$\sin C = \frac{1}{2}$Thus,$C = 30^{\circ}$ or $C = 150^{\circ}$.Checking the original equations with $C = 150^{\circ}$ leads to a contradiction,so $C = 30^{\circ}$.

In $\triangle ABC$,we are given that $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$. Then,the measure of $\angle C$ is $....^{\circ}$

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