Doubling the angle from the ladder

For the first angle $x$ , $\sin x = \frac{15}{25} = \frac{3}{5}$ and $\cos x = \frac{20}{25} = \frac{4}{5}$ .

For the second angle $2x$ , $H' = 25 \sin 2x = 25 \cdot 2 \sin x \cos x = 25 \cdot 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = 24$ , and $D' = 25 \cos 2x = 25 (\cos^2 x - \sin^2 x) = 25 ((\frac{4}{5})^2 - (\frac{3}{5})^2) = 7$ .

Therefore, $H' + D' = 24 + 7 = \boxed{31}$ .

Since the information is given that all lengths are integers again, we just have to find another pythagorean triple with a hypothenuse of 25. The only possibility is $(7, 24, 25)$ , so the answer is $7+24 = \boxed{31}$ .

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$({\color{#D61F06}7}, 24, 25)$ can be generated by noting that $25 = \frac {{\color{#D61F06}7}^2+1}{2}$ . Then, the second leg is $\frac {{\color{#D61F06}7}^2-1}{2} = 24$ .

$\tan (x) = \dfrac {15}{20} = \dfrac 34$

Using trigonometric formula $\tan(2β) = \dfrac {2\tan (β)}{1 - \tan^2(β)}$

$\tan (2x) = \dfrac {2\times \frac 34}{1 - \frac {9}{16}} = \dfrac {24}{7}$

Therefore, $H'= 24k , D'= 7k$

Now, $(24k)^2 + (7k)^2 = (25)^2 \Rightarrow 625 k^2 = 625 \Rightarrow k = 1$

Therefore, $H'+ D'= 24(1) + 7(1) =\boxed{31}$

25^2 = 625 so need to find 2 integers whose squares sum to 625.

Largest must be less than 25.

24^2 = 576.

625 - 576 = 49.

Sqrt. of 49 is 7.

24 + 7 = 31.