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

Answer: (C)

Let $ a=\sin \alpha ,b=\cos \alpha ,c=\sqrt{1+\sin \alpha \cos \alpha } $ then $ \cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab} $ $ \Rightarrow $ $ \cos C=\frac{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +1-\sin \alpha \cos \alpha }{2\sin \alpha \cos \alpha } $ $ \Rightarrow $ $ \cos C=-\frac{\sin \alpha \cos \alpha }{2\sin \alpha \cos \alpha } $ $ \Rightarrow $ $ \cos C=-\frac{1}{2}=\cos 120{}^\circ $ $ \Rightarrow $ $ \angle C=120{}^\circ $