Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100 m, ACB = 30° and ADB = 45°.

AB	= tan 30° =	1	AC = AB x 3 = 1003 m.
AC		3

AB	= tan 45° = 1          AD = AB = 100 m.
AD

CD = (AC + AD)	= (1003 + 100) m
	= 100(3 + 1)
	= (100 x 2.73) m
	= 273 m.

One of AB, AD and CD must have given.

So, the data is inadequate.

Let AB be the wall and BC be the ladder.

Then, ACB = 60° and AC = 4.6 m.

BC	= 2 x AC
	= (2 x 4.6) m
	= 9.2 m.

Let AB be the observer and CD be the tower.

Draw BE  CD.

Then, CE = AB = 1.6 m,

      BE = AC = 20√3 m.

DE =	20√3	m = 20 m.
	√3

 CD = CE + DE = (1.6 + 20) m = 21.6 m.