The lengths of the sides of a triangle are $\sin\alpha$, $\cos\alpha$ and $\sqrt{(1+\sin\alpha\cos\alpha)}$…

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 $$120^o$$.

• The greatest angle is opposite to the greatest side and the third side is obviously the biggest one so no need to check the other two – Vasya Mar 20 at 15:51
• @Vasya how would you what is the greatest side. Everything is in cos and sin... – rash Mar 21 at 0:02
• because it's greater than $1$ (sine and cosine are positive) – Vasya Mar 21 at 2:44

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}
Here $$1+\sin\alpha\cos\alpha>1>\cos^2\alpha,\sin^2\alpha$$
So, $$\cos A=\dfrac{\cos^2\alpha+\sin^2\alpha-(1+\cos\alpha\sin\alpha)}{2\cos\alpha\sin\alpha}$$
Obviously $$\sqrt{1+sin\alpha cos\alpha}$$ will always be the longest side (always >1).Now we find $$\alpha$$ for which this will be maximum.
$$\frac{d}{d\alpha}\sqrt{1+sin\alpha cos\alpha}=\frac{cos(2\alpha)}{8\sqrt{1+sin\alpha cos\alpha}}$$Equating it to 0 , we have $$\alpha=\frac{\pi}{4}$$. As a check , you can also verify that at $$\alpha=\frac{\pi}{4} ,sin\alpha+cos\alpha>\sqrt{1+sin\alpha cos\alpha}$$ ,it is a triangle surely.
By the marking scheme in this diagram $$sinA=\frac{\sqrt3}{2}$$ which you can easily procure by substituting $$\alpha=\frac{\pi}{4}$$ . This gives $$A=60^\circ ,2A=120^\circ$$ ,which is the angle opposite to the longest side and hence is the largest angle.