Pls give solution of it (maths problem)

The answer is :- ( square root of ) 3 km.
In euclidean geometry the shortest distance between two points is a straight line. So consider a circle of radius 1 with center at point Q, by question B and C are two opposite points of a diameter of the circle ( i.e. points B and C are two extremities ). The point A is a point on the circle such that the length of the straight line AB is 1 , connect A , Q so that AQ=1 since AQ is a radius of the circle.Also , BQ=1 , hence in triangle ABQ , AB=AQ=BQ=1 i.e. triangle ABQ is equilateral , so (angle) <AQB = 60 (degree). Now connect A,C .
CQ = 1 ( for being a radius of the circle). Since B,Q,C are collinear ,
( angle)<BQC = 180 . So (angle)<AQC= <BQC - <AQB = 180 - 60 = 120 .
So, in triangle AQC , AQ=CQ=1 , <AQC=120 , so triangle AQC is isosceles with , <CAQ = <QAC = ( 180 - <AQC ) / 2 = (180 - 120 ) / 2 = 60 / 2 = 30 .
Sine Law gives AC / sin<AQC = CQ / sin <CAQ i.e. AC / sin 120 = 1 / sin30
i.e. 2 AC / (sqrt3) = 2 i.e. AC = square root of 3
= shortest distance between A and C

Because B to C is 2km, you know they are on opposite sides of the circle. That means for any point A, you are going to have a right triangle. Since a^2 + b^2 = c^2 for a right triangle, AC^2 = BC^2 - AB^2 or AC^2 = 2^2 - 1^2 = 3 so AC = sqrt 3.