In triangle ABC, $\angle ABC = 90^\circ$, $BX = 2XC$ and $\sin \angle AXC = \dfrac{3}{5}$. Express as a fraction in its lowest terms,

1. Use the given sine ratio to find the corresponding cosine ratio for $\angle AXC$.

Since $\sin \angle AXC=\dfrac35$, take opposite $=3$ and hypotenuse $=5$.

By Pythagoras, adjacent $=\sqrt{5^2-3^2}=4$.

So
$\tan \angle AXC=\frac{3}{4}.$

2. Relate $\angle AXB$ to $\angle AXC$ and determine the tangent value.

Since $B, X, C$ are on a straight line,
$\angle AXB=180^\circ-\angle AXC.$

Therefore,
$\tan \angle AXB=-\tan \angle AXC=-\frac34.$

1. Find the ratio of the vertical height to $XC$ using $\sin \angle AXC=\dfrac{3}{5}$.

Since $\sin \angle AXC=\dfrac{3}{5}$, the corresponding right-triangle ratio is $3:4:5$.

Thus the vertical height $AB$ is proportional to $3$ parts and $XC$ is proportional to $4$ parts, so

$\frac{AB}{XC}=\frac{3}{4}.$

2. Use $BX=2XC$ to find $\tan \angle ACB$.

Given $BX=2XC$, the whole base is

$BC=BX+XC=2XC+XC=3XC.$

In right triangle $ABC$,

$\tan \angle ACB=\frac{AB}{BC}=\frac{\frac{3}{4}XC}{3XC}=\frac{1}{4}.$