If sin A = -frac{24}{25},cos B = frac{15}{17} | vedclass

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(C) Given $\sin A = -\frac{24}{25}$. Since $A$ is not in the $4^{	ext{th}}$ quadrant and $\sin A < 0$,$A$ must be in the $3^{	ext{rd}}$ quadrant. Th

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$3^{	ext{rd}}$ quadrant

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(C) Given $\sin A = -\frac{24}{25}$. Since $A$ is not in the $4^{	ext{th}}$ quadrant and $\sin A Given $\cos B = \frac{15}{17}$. Since $B$ is not in the $1^{	ext{st}}$ quadrant and $\cos B > 0$,$B$ must be in the $4^{	ext{th}}$ quadrant. Thus,$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (\frac{15}{17})^2} = -\sqrt{1 - \frac{225}{289}} = -\sqrt{\frac{64}{289}} = -\frac{8}{17}$.Now,$\sin(A+B) = \sin A \cos B + \cos A \sin B = (-\frac{24}{25})(\frac{15}{17}) + (-\frac{7}{25})(-\frac{8}{17}) = -\frac{360}{425} + \frac{56}{425} = -\frac{304}{425} $\cos(A+B) = \cos A \cos B - \sin A \sin B = (-\frac{7}{25})(\frac{15}{17}) - (-\frac{24}{25})(-\frac{8}{17}) = -\frac{105}{425} - \frac{192}{425} = -\frac{297}{425} Since both $\sin(A+B) < 0$ and $\cos(A+B) < 0$,$(A+B)$ lies in the $3^{	ext{rd}}$ quadrant.

If $\sin A = -\frac{24}{25}$,$\cos B = \frac{15}{17}$,$A$ does not belong to the $4^{\text{th}}$ quadrant,and $B$ does not belong to the $1^{\text{st}}$ quadrant,then $(A+B)$ lies in which quadrant?

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