If in a triangle ABC,cos A cos B + sin A sin | vedclass

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Question

(A) We have,$\cos A \cos B + \sin A \sin B \sin C = 1$.From the given relation,$\sin C = \frac{1 - \cos A \cos B}{\sin A \sin B} \leq 1$.This implies

Vedclass · Accepted Answer

$1 : 1 : \sqrt{2}$

Vedclass · Answer

(A) We have,$\cos A \cos B + \sin A \sin B \sin C = 1$.From the given relation,$\sin C = \frac{1 - \cos A \cos B}{\sin A \sin B} \leq 1$.This implies $1 - \cos A \cos B \leq \sin A \sin B$,which simplifies to $1 \leq \cos A \cos B + \sin A \sin B$.Thus,$1 \leq \cos(A - B)$. Since $\cos(A - B) \leq 1$,we must have $\cos(A - B) = 1$,which implies $A - B = 0$,or $A = B$.Substituting $A = B$ into the original equation,we get $\cos^2 A + \sin^2 A \sin C = 1$.$\sin^2 A \sin C = 1 - \cos^2 A = \sin^2 A$.Since $A$ is an angle of a triangle,$\sin A 
eq 0$,so $\sin C = 1$,which means $C = 90^{\circ}$.Since $A + B + C = 180^{\circ}$ and $A = B$,we have $2A + 90^{\circ} = 180^{\circ}$,so $A = B = 45^{\circ}$.By the Sine Law,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.$\frac{a}{\sin 45^{\circ}} = \frac{b}{\sin 45^{\circ}} = \frac{c}{\sin 90^{\circ}}$.$\frac{a}{1/\sqrt{2}} = \frac{b}{1/\sqrt{2}} = \frac{c}{1}$.Therefore,$a : b : c = 1 : 1 : \sqrt{2}$.

If in a triangle $ABC$,$\cos A \cos B + \sin A \sin B \sin C = 1$,then $a : b : c =$

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