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Connecting CE, CF⊥AB, vertical foot f,

CE is the hypotenuse midline, CE=AB/2=5,

CE=BE=AE,EBC = ECB,

< FEC = 2 < b = 30 degrees, (the outer angle is equal to the sum of two non-adjacent inner angles),

CF=CE/2=5/2, (the opposite side of 30 degrees is half of the hypotenuse),

EF=√3CF=5√3/2,

AF=AE-EF=5-5√3/2

According to Pythagorean theorem,

AC^2=AF^2+CF^2,

AC=√(50-50√3)

=5√(2-√3)

=5√[(8-4√3)/4]

=5√[(6-2√ 12+2)/4]

=5√[(√6-√2)^2/4]

=5(√6-√2)/2.