# If the side lengths of a triangle increase and the third side is fixed, the opposite angle decreases

Suppose that I have an Euclidean triangle in the plane with side lengths $$a,b,c$$. Denote the angle opposite $$c$$ by $$\theta$$.

I am trying to prove, that if we keep $$c$$ fixed, and increase $$a,b$$, then $$\theta$$ should get smaller.

That is, if $$\tilde a,\tilde b,c$$ are side lengths of another triangle, and $$\tilde \theta$$ is the corresponding angle, then

$$\tilde a \ge a, \, \tilde b \ge b \Rightarrow \tilde \theta \le \theta.$$

I tried to prove this via the law of cosines:

$$\cos(\theta)= \frac{a^2+b^2-c^2}{2ab}\le \frac{\tilde a^2+\tilde b^2-c^2}{2\tilde a \tilde b}=\cos(\tilde \theta),$$

but somehow got stucked.

Is there a simple algebraic proof of this inequality? or alternatively, geometric proof?

This is a simple counterexample: • Thanks. I also realized the statement doesn't always hold. But it does hold whenever all the angles are less than $\frac{\pi}{2}$ (by the sine law). And perhaps also when $\theta$ itself is bigger than $\frac{\pi}{2}$. BTW, may I ask how did you produce this example? – Asaf Shachar Apr 22 at 15:49
• @Asaf Shachar: this example was prepared using Asymptote, "Asymptote is a powerful descriptive vector graphics language that provides a natural coordinate-based framework for technical drawing. Labels and equations are typeset with LaTeX, for high-quality PostScript output." Also, you can check out this forum. – g.kov Apr 22 at 15:55

The way I understood the given problem which to me makes sense:

"If we keep positions of $$(A,B)$$ fixed and at least one of lengths of $$(a,b)$$ increased, then show that $$\theta$$ included between them decreases." An ellipse is a convenient choice since adjacent legs sum up to a constant. Since confocal ellipses do not intersect it suffices to choose on layer as shown: $$h= a \sin \beta=b \sin \alpha\,;\tag1$$ Lawof Sines $$\frac{\sin \theta}{c}=\frac{\sin \alpha}{a}=\frac{\sin \beta}{b}=\frac{\sin \alpha+\sin \beta}{a+b}=\frac{\sin \alpha+\sin \beta}{p}\tag2$$ where $$p$$ is major axis $$>c$$.

Since $$\beta=\alpha-\theta$$ $$\frac{\sin (\theta-\alpha) +\sin \beta}{\sin \theta}=\frac{p}{c}>1 \tag3$$

$$\dfrac{p}{c}$$ will be always greater than $$1$$ and expression at left increases monotonously with $$\theta$$, proving the proposition... the way I understood it.