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

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Question

(c) $a = \sin \alpha ,\,b = \cos \alpha ,\,c = \sqrt {1 + \sin \alpha \cos \alpha } $
$\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$=$\frac{{{{\sin

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(c) $a = \sin \alpha ,\,b = \cos \alpha ,\,c = \sqrt {1 + \sin \alpha \cos \alpha } $
$\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$=$\frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha - (1 + \sin \alpha .\cos \alpha )}}{{2\sin \alpha .\cos \alpha }}$
= $\frac{{1 - 1 - \sin \alpha \cos \alpha }}{{2\sin \alpha \cos \alpha }}$
$\cos C = \frac{{ - 1}}{2} = \cos \frac{{2\pi }}{3}$==> $C = \frac{{2\pi }}{3} = 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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