If cot A=frac{11}{60},cos B=frac{7}{25} and n | vedclass

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(A) Given,$\cot A=\frac{11}{60}$. Since $A$ is not in the first quadrant and $\cot A > 0$,$A$ must be in the third quadrant $(Q_3)$. Thus,$	an A = \f

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(A) Given,$\cot A=\frac{11}{60}$. Since $A$ is not in the first quadrant and $\cot A > 0$,$A$ must be in the third quadrant $(Q_3)$. Thus,$	an A = \frac{60}{11}$.Given $\cos B = \frac{7}{25}$. Since $B$ is not in the first quadrant and $\cos B > 0$,$B$ must be in the fourth quadrant $(Q_4)$.In $Q_4$,$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (\frac{7}{25})^2} = -\frac{24}{25}$.Thus,$	an B = \frac{\sin B}{\cos B} = \frac{-24/25}{7/25} = -\frac{24}{7}$.Using the formula $	an B = \frac{2 	an(B/2)}{1 - 	an^2(B/2)}$,we have $-\frac{24}{7} = \frac{2 	an(B/2)}{1 - 	an^2(B/2)}$.Let $t = 	an(B/2)$. Then $-\frac{24}{7} = \frac{2t}{1-t^2}$ $\Rightarrow -24 + 24t^2 = 14t$ $\Rightarrow 12t^2 - 7t - 12 = 0$.Solving the quadratic equation: $12t^2 - 16t + 9t - 12 = 0$ $\Rightarrow 4t(3t - 4) + 3(3t - 4) = 0$ $\Rightarrow (4t + 3)(3t - 4) = 0$.So,$t = \frac{4}{3}$ or $t = -\frac{3}{4}$.Since $B \in Q_4$,$\frac{3\pi}{2} Now,$	an(A + B/2) = \frac{	an A + 	an(B/2)}{1 - 	an A \cdot 	an(B/2)} = \frac{\frac{60}{11} - \frac{3}{4}}{1 - (\frac{60}{11})(-\frac{3}{4})} = \frac{\frac{240 - 33}{44}}{1 + \frac{180}{44}} = \frac{207/44}{224/44} = \frac{207}{224}$.Since $	an(A + B/2) > 0$,the angle $(A + B/2)$ lies in the first quadrant $(Q_1)$.

If $\cot A=\frac{11}{60}$,$\cos B=\frac{7}{25}$ and neither $A$ nor $B$ is in the first quadrant,then $\left(A+\frac{B}{2}\right)$ lies in the quadrant

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