Question

The side 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

• A. ${120}^0$
• B. ${90}^0$
• C. ${60}^0$
• D. ${150}^0$

Answer 1

The correct option is A ${120}^0$

$a=sin\theta$
$b=cos\theta$
And
$c=\sqrt{1+sin\theta .cos\theta }$
Hence
$cosC=\frac{a^2+b^2-c^2}{2ab}$
$=\frac{co{s}^2\theta +si{n}^2\theta -1-sin\theta .cos\theta }{2sin\theta .cos\theta }$
Or
$cosC=\frac{-1}{2}$.
Hence
$c=co{s}^{-1}\left(\frac{-1}{2}\right)$
$=\pi -\frac{\pi }{3}$
Or
$c=\frac{2\pi }{3}$ or ${120}^o$