The sides of a triangle are sin⁡α,cos⁡α and 1+sin⁡αcos⁡α for some α,0<α<π2.Then the greatest angle of the triangle is

a=1+sinαcosα,b=sinα,c=cosα are sides of the trianglecos⁡A=b2+c2-a22bc=sin2⁡α+cos2⁡α−1−sin⁡αcos⁡α2sin⁡αcos⁡α=−12⇒A=120∘.

The sides of a triangle are sin⁡α,cos⁡α and 1+sin⁡αcos⁡α for some α,0<α<π2.Then the greatest angle of the triangle is

a=1+sinαcosα,b=sinα,c=cosα are sides of the trianglecos⁡A=b2+c2-a22bc=sin2⁡α+cos2⁡α−1−sin⁡αcos⁡α2sin⁡αcos⁡α=−12⇒A=120∘.

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The sides of a triangle are $\sin α, \cos α and \sqrt{1 + \sin αcos α}$ for some $α, 0 < α < \frac{π}{2}$ .Then the greatest angle of the triangle is

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90°

60°

120°

150°

answer is C.

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Detailed Solution

$\begin{array}{l} a = \sqrt{1 + \sin α \cos α}, b = \sin α, c = \cos α are sides of the triangle \\ \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{\sin^{2} α + \cos^{2} α - 1 - \sin αcos α}{2 \sin αcos α} = - \frac{1}{2} \\ \Rightarrow A = 120^{\circ} . \end{array}$