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

Question

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 1

Clearly the greatest angle is opposite to the greatest side. Use Cosine Rule to get $$\begin{aligned}(1+\sin \alpha\cos\alpha)&=\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha\cos x\\ \dfrac{\sin\alpha\cos\alpha+1-1}{-2\sin\alpha\cos\alpha}&=\cos x\\ \cos x&=\dfrac{-1}{2}\implies x=\dfrac{2\pi}{3}=120^{\circ}\end{aligned}$$

Answer 2

Hint:

Here $1+\sin\alpha\cos\alpha>1>\cos^2\alpha,\sin^2\alpha$

So, $$\cos A=\dfrac{\cos^2\alpha+\sin^2\alpha-(1+\cos\alpha\sin\alpha)}{2\cos\alpha\sin\alpha}$$

Answer 3

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.