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

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Question

(A) Given the relation $\cos A \cos B + \sin A \sin B \sin C = 1.$Rearranging for $\sin C,$ we get $\sin C = \frac{1 - \cos A \cos B}{\sin A \sin B}.$

Vedclass · Accepted Answer

$1 : 1 : \sqrt{2}$

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(A) Given the relation $\cos A \cos B + \sin A \sin B \sin C = 1.$Rearranging for $\sin C,$ we get $\sin C = \frac{1 - \cos A \cos B}{\sin A \sin B}.$Since $\sin C \le 1,$ we have $\frac{1 - \cos A \cos B}{\sin A \sin B} \le 1.$This implies $1 - \cos A \cos B \le \sin A \sin B,$ or $1 \le \cos A \cos B + \sin A \sin B.$Using the cosine difference formula,$1 \le \cos(A - B).$Since the maximum value of $\cos 	heta$ is $1,$ we must have $\cos(A - B) = 1,$ which implies $A - B = 0$ or $A = B.$Substituting $A = B$ into the original equation,$\sin C = \frac{1 - \cos^2 A}{\sin^2 A} = \frac{\sin^2 A}{\sin^2 A} = 1.$Thus,$C = 90^{\circ}.$ Since $A + B + C = 180^{\circ}$ and $A = B,$ we get $2A = 90^{\circ},$ so $A = B = 45^{\circ}.$Using the sine rule,the sides are proportional to $\sin A : \sin B : \sin C = \sin 45^{\circ} : \sin 45^{\circ} : \sin 90^{\circ} = \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1 = 1 : 1 : \sqrt{2}.$

If in a triangle $ABC,$ $\cos A \cos B + \sin A \sin B \sin C = 1,$ then the sides are proportional to

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