Geometry

             A    E

B                 D                  C  15

      10                  10   

EC / sin 90 = DC / sin75

EC = 10/sin75  

BD = DC

ED = ED

So....by LL,  right triangle BED  congruent to right triangle CED

So BE  = 10/sin75 ≈ 10.35

I just want to add on that we can represent the answer $10\sin(75) =  10\tan(15) = 10\tan(\frac{30}{2}).$
Now we can use the trigonometric half-angle identity $\tan(\frac{\alpha}{2}) = \frac{\sin(\alpha)}{\cos(\alpha)+1}$.
We can input our normal values: $10\tan(\frac{30}{2}) = 10(\tan(\frac{30}{2})) = 10(\frac{\sin(30)}{\cos(30)+1}) = 10(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}+1) = 10(\frac{1}{\sqrt{3}+2}) = 10(2-\sqrt{3})= 20 - 10\sqrt{3}.$
Therefore $BE = \sqrt{(20 - 10\sqrt{3})^2+10^2} = \sqrt{400 + 300 - 400\sqrt{3} + 100} = \sqrt{800 - 400\sqrt{3}} = \sqrt{400(2-\sqrt{3})} = 20\sqrt{2-\sqrt{3}}$