Question

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

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

Answer 1

The correct option is C ${120}^{\circ }$

Let $a=\sin \alpha ,b=\cos \alpha$ and $c=\sqrt{1+\sin \alpha \cos \alpha }$
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We know that in a triangle the side which is largest opposite of that angle is greatest.

In this case $\angle C$ will be greatest.

$\cos C=\frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}\Rightarrow \angle C={120}^{\circ }$