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°
• B. 90°
• C. 120°
• D. 150°

Answer 1

The correct option is C

120°

Let sides are $a=sin\alpha cos\alpha ,c=\sqrt{1+sin\alpha cos\alpha },Then,cosC=\frac{a^2+b^2-c^2}{2ab}(\because \angle )\text{C is the greatest angle}\Rightarrow cosC=\frac{si{n}^2\alpha +co{s}^2\alpha -(1+sin\alpha cos\alpha )}{2sin\alpha cos\alpha }\Rightarrow cosC=\frac{-sin\alpha cos\alpha }{2sin\alpha cos\alpha }\Rightarrow cosC=-\frac{1}{2}=cos{120}^2\Rightarrow \angle C={120}^0$