The sides of a triangle are sin theta, cos th | vedclass

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(B) Let the sides be $a = \sin 	heta$,$b = \cos 	heta$,and $c = \sqrt{1 + \sin 	heta \cos 	heta}$.Since $0 < 	heta < \frac{\pi}{2}$,both $\sin

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$\frac{2 \pi}{3}$

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(B) Let the sides be $a = \sin 	heta$,$b = \cos 	heta$,and $c = \sqrt{1 + \sin 	heta \cos 	heta}$.Since $0 Comparing the squares of the sides: $a^2 = \sin^2 	heta$,$b^2 = \cos^2 	heta$,and $c^2 = 1 + \sin 	heta \cos 	heta$.Since $c^2 = \sin^2 	heta + \cos^2 	heta + \sin 	heta \cos 	heta$,it is clear that $c^2 > a^2$ and $c^2 > b^2$,so $c$ is the longest side.The greatest angle $C$ is opposite to side $c$.Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.$\cos C = \frac{\sin^2 	heta + \cos^2 	heta - (1 + \sin 	heta \cos 	heta)}{2 \sin 	heta \cos 	heta}$.$\cos C = \frac{1 - 1 - \sin 	heta \cos 	heta}{2 \sin 	heta \cos 	heta} = \frac{-\sin 	heta \cos 	heta}{2 \sin 	heta \cos 	heta} = -\frac{1}{2}$.Since $\cos C = -\frac{1}{2}$,$C = 120^{\circ} = \frac{2 \pi}{3}$.

The sides of a triangle are $\sin \theta, \cos \theta$ and $\sqrt{1 + \sin \theta \cos \theta}$ for some $0 < \theta < \frac{\pi}{2}$. Then the greatest angle of the triangle is:

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