The sides of a triangle are $\sin heta, \cos heta$ and $\sqrt{1+\sin heta \cos heta}$ for some $0 \lt heta \lt \frac{\pi}{2}$, then the greatest angle of a triangle is

Question

The sides of a triangle are $\sin 	heta, \cos 	heta$ and $\sqrt{1+\sin 	heta \cos 	heta}$ for some $0 \lt 	heta \lt \frac{\pi}{2}$, then the greatest angle of a triangle is

Aicharya · Accepted Answer

$\mathrm{a}=\sin 	heta, \mathrm{b}=\cos 	heta 	ext { and } \mathrm{c}=\sqrt{1+\sin 	heta \cos 	heta}$ Since $\sqrt{1+\sin 	heta \cos 	heta}$ is greater than $\sin 	heta$ and $\cos 	heta$. $	herefore \quad C$ is the greatest angle, $\begin{aligned} & 	herefore \quad \cos C \ & 	herefore \quad \ & \quad=\frac{a^2+b^2-c^2}{2 a b} \ & \ & \quad=-\frac{1}{2}=\cos 120^{\circ} \ & 	herefore \quad C \end{aligned}$

The sides of a triangle are $\sin \theta, \cos \theta$ and $\sqrt{1+\sin \theta \cos \theta}$ for some $0…

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