Solution:

Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
∠BAD = 45°, ∠BCD = 60°
$$\eqalign{
& \tan {45^ \circ } = \frac{{BD}}{{BA}} \cr
& \Rightarrow 1 = \frac{{60}}{{BA}} \cr
& \Rightarrow BA = 60\,m\,........\left( {\text{i}} \right) \cr
& \tan {60^ \circ } = \frac{{BD}}{{BC}} \cr
& \Rightarrow \sqrt 3 = \frac{{60}}{{BC}} \cr
& \Rightarrow BC = \frac{{60}}{{\sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60\sqrt 3 }}{3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\sqrt 3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20 \times 1.73 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 34.6\,m\,..........\left( {{\text{ii}}} \right) \cr} $$
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (i) and (ii)]
= 94.6 m