Question

Please help, Brainliest if correct!!!

Answer 1

See the attached pic below

In the triangle formed by the statue and the boat, we have

$\tan23^\circ=\dfrac x{110+y}$

and

$\tan38^\circ50'=\dfrac xy$

We can solve either equation for $y$, then substitute that into the other equation. First, let's abbreviate $t_1=\tan23^\circ$ and $t_2=\tan38^\circ50'$. Then

$y=\dfrac x{t_2}$

$\implies t_1=\dfrac x{110+\frac x{t_2}}$

$\implies\dfrac{t_1}{t_2}=\dfrac x{110t_2+x}$

$\implies(110t_2+x)\dfrac{t_1}{t_2}=x$

$\implies110t_1=\left(1-\dfrac{t_1}{t_2}\right)x$

$\implies x=\dfrac{110t_1}{1-\frac{t_1}{t_2}}=\dfrac{110t_1t_2}{t_2-t_1}$

$\implies\boxed{x\approx99}$