It's another Pythagoras' triplet? No way!

Figure 1. Starting right triangle of Pythagorean triple $3$ - $4$ - $5$ . The red angle is indicated by $\theta$ . Let $\theta$ denote the shortest angle of the right triangle. Since the shortest angle must "face" the shortest side of the triangle, and the shortest side length is $3$ , $\theta$ must "face" the side with the length of $3$ .

Note: Another way to see this is to consider the Law of Sine : $\dfrac{A}{\sin(\alpha)} = \dfrac{B}{\sin(\beta)} = \dfrac{G}{\sin(\gamma)}$ where $A, B$ and $G$ are side lengths that are opposite to the angles $\alpha, \beta$ and $\gamma$ respectively. Since the ratio has the direct relationship, it is clear that as you increase the side length of the triangle, the corresponding angle increases. Likewise, decreasing the side length of the triangle decreases the angle.

Figure 2. New right triangle Let's determine $2\theta$ by trigonometry. Since from Figure 1 $\theta = \arctan\left(\dfrac{3}{4}\right)$ , we have $2\theta = 2\arctan\left(\dfrac{3}{4}\right)$ . Applying the following identity (see Footnote ), $2\arctan(\alpha) = \arctan\left(\dfrac{2\alpha}{1 - \alpha^2}\right)$ where $\alpha = \dfrac{3}{4}$ , gives $\begin{array}{rl} 2\arctan\left(\dfrac{3}{4}\right) &= \arctan\left(\dfrac{2\cdot \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2}\right)\\ &= \arctan\left(\dfrac{24}{7}\right) \end{array}$ which shows that the new opposite side length is $24$ , and the new adjacent side length is $7$ . Thus, we obtain the new Pythagorean triple $\boxed{7:24:25}$ .

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Footnote

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To show that $2\arctan(\alpha) = \arctan\left(\dfrac{2\alpha}{1 - \alpha^2}\right)$ express $2\arctan(\alpha)$ as $2\arctan(\alpha) = \arctan(\alpha) + \arctan(\alpha)$ Since $\arctan(x) + \arctan(y) = \arctan\left(\dfrac{x + y}{1 - xy}\right)$ this shows that $2\arctan(\alpha) = \arctan\left(\dfrac{\alpha + \alpha}{1 - \alpha \cdot \alpha}\right) = \arctan\left(\dfrac{2\alpha}{1 - \alpha^2}\right)$

In a 3:4:5 triangle, the smallest angle faces the smallest side, or 3. Call this angle x.

Thus, sin(x) = 0.6, and cos(x) = 0.8

If the angle becomes 2x, we can find the ratio of the opposite side to the hypotenuse, or sin(2x).

Using sin(2x) = 2cos(x)sin(x), we plug in the values we previously found and get sin(2x) = 0.96

This means that the ratio of the side opposite to 2x to the hypotenuse is 0.96:1 = 96:100 = 24:25.

Using the Pythagorean theorem, we can find the third side: 7.

Thus, the new ratio is 7:24:25

Firstly we assume that the sides of triangle be 3k, 4k , and 5k for some non zero positive number k . We know that angle opposite to smallest side is smallest. Thus $\theta$ is the angle opposite to the side 3k . Hence, From cosine rule we write, $\cos\theta$ = $\frac{{5k}^{2} +{4k}^{2}-{3k}^{2}}{2\times (4k)\times (5k)}$ = $\frac{4}{5}$ . Now using trigonometry for multiple angles, we write $\cos 2\theta$ =2 $\cos^2 \theta$ -1 = $\frac{7}{25}$ and we get $\sin2\theta$ = $\frac{24}{25}$ $\Big($ $\sin2\theta$ $\neq$ $\frac{-24}{25}$ because $2\theta$ is less than $\pi$ $\Big)$ . Since the triangle is right angled , we get the sides of the new triangle as 7m , 24m , 25m for some non zero positive number 'm' . Thus we get the ration of sides as $7:24:25$