# What term/identity/theorem states that given a triangle the largest angle

What term/identity/theorem states that given a triangle the largest angle must be opposite the longest side?
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Calvin Oneill
Step 1
It is a result of the law of sines:
$\frac{\mathrm{sin}A}{a}=\frac{\mathrm{sin}B}{b}=\frac{\mathrm{sin}C}{c}$
This means tha the largest of a,b,c corresponds to the largest of $\mathrm{sin}A,\mathrm{sin}B,\mathrm{sin}C.$ .
Assume A is the largest angle.
If $A\le \frac{\pi }{2},$ , then, since sinx is increasing on $\left[0,\pi /2\right],$ , $\mathrm{sin}A$ is the largest of the sines, so a is largest.
If $A>\frac{\pi }{2},$ , then $\mathrm{sin}\left(A\right)=\mathrm{sin}\left(\pi -\left(B+C\right)\right)=\mathrm{sin}\left(B+C\right),$ , with $0\le B,C\le B+C<\frac{\pi }{2}.$. So $\mathrm{sin}A\ge \mathrm{sin}B,\mathrm{sin}C,$ and again sinA is the largest value, and hence a is the longest length.
Step 2
Alternatively, more geometrically, find the circle which goes through A, B, C, and then, at least for A, B, C acute angles, the lengths of the sides and the lengths of the corresponding arc lengths, and the arc lengths correspond to twice the opposite angle.
You then have to deal with the obtuse case separately. Not hard to do - type get the length corresponds to the arc of $2\pi -2A=2B+2C>2B,2C$ and thus a is still longer then both b, c.
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revistasbibxjm87
Such result is known as:
"The side of a triangle opposite the largest angle is the longest side."
You can find its proof in any plane Euclidean geometry book.