In triangle ABC,if A = tan^{-1} 2 and B = tan | vedclass

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(B) In any triangle $ABC$,the sum of angles is $A + B + C = 180^{\circ}$.Given $A = 	an^{-1} 2$ and $B = 	an^{-1} 3$.We know that $	an(A + B) = \fr

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$45$

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(B) In any triangle $ABC$,the sum of angles is $A + B + C = 180^{\circ}$.Given $A = 	an^{-1} 2$ and $B = 	an^{-1} 3$.We know that $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Substituting the values,$	an(A + B) = \frac{2 + 3}{1 - (2)(3)} = \frac{5}{1 - 6} = \frac{5}{-5} = -1$.Since $A$ and $B$ are angles of a triangle and $	an A = 2, 	an B = 3$ (both positive),$A$ and $B$ are acute angles.Thus,$A + B$ must be in the second quadrant because $	an(A + B) = -1$.Therefore,$A + B = 180^{\circ} - 45^{\circ} = 135^{\circ}$.Since $A + B + C = 180^{\circ}$,we have $135^{\circ} + C = 180^{\circ}$.Hence,$C = 180^{\circ} - 135^{\circ} = 45^{\circ}$.

In $\triangle ABC$,if $A = \tan^{-1} 2$ and $B = \tan^{-1} 3$,then $C = $ . . . . . . (in $^{\circ}$)

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