Answer :
Let the ships be at A and B and the top of the light house be C as shown in the figure.
The distance between the ships id AB
AB = AO+OB
given that OC = 200m
tan45 = OC/OB
⇒ 1 = 200/OB
⇒OB = 200m
tan 60 = OC/AO
⇒√3 = 200/AO
⇒AO = 200/√3 = 115.47m
AB = AO+OB = 200+115.47 = 315.47m
Distance between ships is 315.47m
The distance between the ships id AB
AB = AO+OB
given that OC = 200m
tan45 = OC/OB
⇒ 1 = 200/OB
⇒OB = 200m
tan 60 = OC/AO
⇒√3 = 200/AO
⇒AO = 200/√3 = 115.47m
AB = AO+OB = 200+115.47 = 315.47m
Distance between ships is 315.47m
Let P and Q be the two ships and AB be the light house
tan45 =AB/BP
1=200/BP
BP=200
tan60=AB/BQ
1.732=200/ BQ
BQ=200/1.732
BQ=115.7
PQ=PB+BQ
PQ=200+115.7
PQ=315.7