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

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(B) Given the equation $\cos A \cos B + \sin A \sin B \sin C = 1$ in $	riangle ABC$. Since $\sin C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin

Vedclass · Accepted Answer

$1+\sqrt{2}$

Vedclass · Answer

(B) Given the equation $\cos A \cos B + \sin A \sin B \sin C = 1$ in $	riangle ABC$. Since $\sin C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin C \le \cos A \cos B + \sin A \sin B = \cos(A - B)$. Since $\cos(A - B) \le 1$,the equality holds only if $\cos(A - B) = 1$ and $\sin C = 1$. This implies $A = B$ and $C = 90^{\circ}$. Since $A + B + C = 180^{\circ}$,we have $2A + 90^{\circ} = 180^{\circ}$,so $A = 45^{\circ}$ and $B = 45^{\circ}$. Thus,$\sin A + \sin B + \sin C = \sin 45^{\circ} + \sin 45^{\circ} + \sin 90^{\circ} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1 = \sqrt{2} + 1$.

In a 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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