In triangle ABC,if cos A cos B + sin A sin B | vedclass

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(B) Given $\cos A \cos B + \sin A \sin B \sin C = 1$. Since $\sin C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin C \le \cos A \cos B + \sin A \s

Vedclass · Accepted Answer

$1+\sqrt{2}$

Vedclass · Answer

(B) Given $\cos A \cos B + \sin A \sin B \sin C = 1$. Since $\sin C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin C \le \cos A \cos B + \sin A \sin B = \cos(A-B)$. Thus,$\cos(A-B) \ge 1$. Since the maximum value of $\cos(A-B)$ is $1$,we must have $\cos(A-B) = 1$,which implies $A = B$. Also,this requires $\sin C = 1$,so $C = 90^\circ$. Since $A+B+C = 180^\circ$ and $A=B$,we have $2A + 90^\circ = 180^\circ$,so $A = B = 45^\circ$. Now,$\sin A + \sin B + \sin C = \sin 45^\circ + \sin 45^\circ + \sin 90^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1 = \frac{2}{\sqrt{2}} + 1 = \sqrt{2} + 1$.

In triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C =$

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