If cos A = -frac{60}{61} and tan B = -frac{7} | vedclass

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(D) Given $\cos A = -\frac{60}{61}$ and $	an B = -\frac{7}{24}$. Since $A$ and $B$ are not in the second quadrant,$A$ must be in the third quadrant (

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$4$

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(D) Given $\cos A = -\frac{60}{61}$ and $	an B = -\frac{7}{24}$. Since $A$ and $B$ are not in the second quadrant,$A$ must be in the third quadrant (as $\cos A For $A \in (\pi, \frac{3\pi}{2})$,$	an A = \frac{11}{60}$.For $B \in (\frac{3\pi}{2}, 2\pi)$,$\frac{B}{2} \in (\frac{3\pi}{4}, \pi)$,which is in the second quadrant.Using $	an B = \frac{2 	an(B/2)}{1 - 	an^2(B/2)} = -\frac{7}{24}$,we get $7 	an^2(B/2) - 48 	an(B/2) - 7 = 0$,which gives $	an(B/2) = 7$ or $	an(B/2) = -1/7$. Since $\frac{B}{2}$ is in the second quadrant,$	an(B/2) = -7$.Now,$	an(A + \frac{B}{2}) = \frac{	an A + 	an(B/2)}{1 - 	an A 	an(B/2)} = \frac{11/60 - 7}{1 + (11/60)(7)} = \frac{11 - 420}{60 + 77} = \frac{-409}{137} Since $A \in (\pi, \frac{3\pi}{2})$ and $\frac{B}{2} \in (\frac{3\pi}{4}, \pi)$,$A + \frac{B}{2} \in (\frac{7\pi}{4}, \frac{5\pi}{2})$,which corresponds to the fourth quadrant.

If $\cos A = -\frac{60}{61}$ and $\tan B = -\frac{7}{24}$ and neither $A$ nor $B$ is in the second quadrant,then the angle $A + \frac{B}{2}$ lies in the quadrant:

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