# 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 Mar 20, 2019 at 15:51
• @Vasya how would you what is the greatest side. Everything is in cos and sin...
– rash
Mar 21, 2019 at 0:02
• because it's greater than $1$ (sine and cosine are positive) Mar 21, 2019 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.