For a certain hyperbola (x^2/a^2)-(y^2/b^2)=1 where a is greater than b, the angle between the asymptotes is 60 degrees. Find a/b
tan (theta / 2) = tan (60/ 2) = tan (30) = 1 /sqrt (3) = b / a
So
tan (60) = sqrt (3) / 1 = a / b
+38 
