4.6. The lengths of the sides of an acute-angled triangle form an arithmetic progression with a common difference of 5 cm. Find the greatest number with the following property: the length of the largest side of any triangle of the specified type is greater than this number.
4.6. Let in $\triangle A B C$ (Fig. 4.5) $A B=x, B C=$ $=x+5$ and $A C=x+10$, where $x>5$. According to the Law of Cosines, we have

$$
(x+10)^{2}=x^{2}+(x+5)^{2}-2 x(x+5) \cos B
$$

from which

$$
\cos B=\frac{x^{2}-10 x+75}{2 x(x+5)}=\frac{(x+5)(x-15)}{2 x(x+5)}=\frac{x-15}{2 x}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-576.jpg?height=405&width=279&top_left_y=436&top_left_x=872)

Fig. 4.5

$$
\text { Since } 0<\cos B<1, \text { then } 0<\frac{x-15}{2 x}<1 \text {. The }
$$

smallest value of $x$ for which this inequality holds is $x=15$. Therefore, the largest number with the property specified in the condition is $x+10=25$.

Answer: 25 cm.