The sides of a triangle are sin alpha,cos alp | vedclass

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

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) Let the sides be $a = \sin \alpha$,$b = \cos \alpha$,and $c = \sqrt{1 + \sin \alpha \cos \alpha}$.Since $0  \sin^2 \alpha + \cos^2 \alpha = 1$.Using the Law of Cosines for the angle $C$ opposite to side $c$:$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$\cos C = \frac{\sin^2 \alpha + \cos^2 \alpha - (1 + \sin \alpha \cos \alpha)}{2 \sin \alpha \cos \alpha}$$\cos C = \frac{1 - 1 - \sin \alpha \cos \alpha}{2 \sin \alpha \cos \alpha}$$\cos C = \frac{-\sin \alpha \cos \alpha}{2 \sin \alpha \cos \alpha} = -\frac{1}{2}$Since $\cos C = -\frac{1}{2}$,the angle $C = 120^\circ$.

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

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