See the attached pic below
In the triangle formed by the statue and the boat, we have
[tex]\tan23^\circ=\dfrac x{110+y}[/tex]
and
[tex]\tan38^\circ50'=\dfrac xy[/tex]
We can solve either equation for [tex]y[/tex], then substitute that into the other equation. First, let's abbreviate [tex]t_1=\tan23^\circ[/tex] and [tex]t_2=\tan38^\circ50'[/tex]. Then
[tex]y=\dfrac x{t_2}[/tex]
[tex]\implies t_1=\dfrac x{110+\frac x{t_2}}[/tex]
[tex]\implies\dfrac{t_1}{t_2}=\dfrac x{110t_2+x}[/tex]
[tex]\implies(110t_2+x)\dfrac{t_1}{t_2}=x[/tex]
[tex]\implies110t_1=\left(1-\dfrac{t_1}{t_2}\right)x[/tex]
[tex]\implies x=\dfrac{110t_1}{1-\frac{t_1}{t_2}}=\dfrac{110t_1t_2}{t_2-t_1}[/tex]
[tex]\implies\boxed{x\approx99}[/tex]
