tan A = frac{-60}{11} and A does not lie in t | vedclass

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(B) Given $	an A = \frac{-60}{11}$. Since $	an A$ is negative,$A$ lies in the $2^{	ext{nd}}$ or $4^{	ext{th}}$ quadrant. Given $A$ does not lie in

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$19$

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(B) Given $	an A = \frac{-60}{11}$. Since $	an A$ is negative,$A$ lies in the $2^{	ext{nd}}$ or $4^{	ext{th}}$ quadrant. Given $A$ does not lie in the $4^{	ext{th}}$ quadrant,$A$ must lie in the $2^{	ext{nd}}$ quadrant.In the $2^{	ext{nd}}$ quadrant,$\operatorname{cosec} A$ is positive. Using the triangle with sides $60, 11, 61$,we get $\operatorname{cosec} A = \frac{61}{60}$.Given $\sec B = \frac{41}{9}$. Since $\sec B$ is positive,$B$ lies in the $1^{	ext{st}}$ or $4^{	ext{th}}$ quadrant. Given $B$ does not lie in the $1^{	ext{st}}$ quadrant,$B$ must lie in the $4^{	ext{th}}$ quadrant.In the $4^{	ext{th}}$ quadrant,$\cot B$ is negative. Using the triangle with sides $40, 9, 41$,we get $\cot B = -\frac{9}{40}$.Now,$K = \operatorname{cosec} A + \cot B = \frac{61}{60} - \frac{9}{40}$.Finding a common denominator $(120)$: $K = \frac{122 - 27}{120} = \frac{95}{120} = \frac{19}{24}$.Therefore,$24K = 24 	imes \frac{19}{24} = 19$.

$\tan A = \frac{-60}{11}$ and $A$ does not lie in the $4^{\text{th}}$ quadrant. $\sec B = \frac{41}{9}$ and $B$ does not lie in the $1^{\text{st}}$ quadrant. If $\operatorname{cosec} A + \cot B = K$,then $24K =$

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