The lengths of the sides of a triangle are $\sin\alpha$, $\cos\alpha$ and $\sqrt{(1+\sin\alpha\cos\alpha)}$, where $0^o < \alpha < 90^o$. The measure of its greatest angle is.......
What I have tried.
By using Cosine Rule,
$$\sqrt{(1+\sin\alpha\cos\alpha)} = \sin^2 \alpha + \cos^\alpha + 2(\sin\alpha)(\cos\alpha)(\cos x)$$
Letting $x$ be an angle for the opposite to $\sqrt{(1+\sin\alpha\cos\alpha)}$,
But my confusion here is how would I know that $x$ is the greatest angle. Do I have to do this step for all other sides? or Is there any shortcut here? or Am I doing it correctly?
The answer is $120^o$.
Answer
Clearly the greatest angle is opposite to the greatest side. Use Cosine Rule to get $$\begin{aligned}(1+\sin \alpha\cos\alpha)&=\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha\cos x\\ \dfrac{\sin\alpha\cos\alpha+1-1}{-2\sin\alpha\cos\alpha}&=\cos x\\ \cos x&=\dfrac{-1}{2}\implies x=\dfrac{2\pi}{3}=120^{\circ}\end{aligned}$$
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