Angles and Ladders

Using the reference image above, (PS I am very good at drawing and that representation is not to scale)

Using left triangle, $sinA = \dfrac{15}{25} = \dfrac{3}{5} \Rightarrow cosA = \sqrt{1-sin^2 A} = \dfrac{4}{5}$

Coming to right triangle, the total angle is 2A (as the initial angle doubles), thus $sin2A = \dfrac{x+15}{25}$

$\rightarrow$ Expanding $sin2A$ as $2sinAcosA$

$2sinAcosA = \dfrac{x+15}{25} \Rightarrow 2 \times \dfrac{4}{5} \times \dfrac{3}{5} = \dfrac{x+15}{25}$

Simplify to get, $\boxed{x=9}$

arcsin(20/25) = 53.13. Double this to get 106.26. 25*sin(106.26) = 24. 24-20 = 4. Incidentally, it's worth noting that he's moved it through vertical so that it's no longer leaning against the wall, which you might miss if you just used a double angle formula.