+743 In triangle $ABC,$ $\angle C = 90^\circ.$ A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$ If the radius of the semicircle is equal to $1$ and $BC = \sqrt{3}$, then find $AB$.
+129885 
Triangle COB is similar to Triangle DOB by HL
tan (angle COB) = 1/sqrt(3) = sqrt (3) / 3 = tan angle DOB
tan angle DBC = tan angle ABC
tan (angle ABC) = tan (angle COB + angle DOB) = tan (2 angle COB) = AC / BC
tan (angle COB + angle DOB ) = [ tan COB + tan DOB ] / [ 1 - tan COB*tan DOB] =
[ 2sqrt (3)/3 ] / [ 1 - (sqrt (3) / 3)^2 ] = (2sqrt (3)/3) / [ 1 - 1/3] = (2sqrt(3)/3) / (2/3) =
sqrt 3 = AC / BC
sqrt 3 = AC / sqrt 3
sqrt 3 * sqrt 3 = AC = 3
