Let the height of light house be AB=100 m and let D be the ship which is sailing towards it. Also let DC = x m.
Here, `angleCAD=60^(@)-30^(@)=30^(@)`
`:. angleCDA=angleCAD" (each "30^(@)`)
`:. DC=AC=x …(1)`
(angles opposite to equal sides are equal)
Now, in right `DeltaABC`, ,brgt `sin60^(@)=(AB)/(AC)rArrsqrt3/2=100/(DC) [from(1)]`
`rArr DC=200/sqrt3=(200sqrt3)/3=(200xx1.73)/3`
`:. DC =346/3=115.3 m`
Hence, distance travelled by ship is 115.3 m.