The answer is
First of all, note that we can solve all angles, since we have
[tex] \hat{A}+\hat{B}+\hat{C} = 180 \iff \hat{A}+90+44.25 = 180 \iff \hat{A} = 180-90-44.25 = 45.75 [/tex]
where [tex] \hat{A} [/tex] is the angle centered in vertex A, and so on.
Now we can use the law of sines, which states that the ratio between a side and the sine of the opposite angles is constant.
So, you would have
[tex] \frac{AC}{\sin(\hat{B})} = \frac{BC}{\sin(\hat{A})} [/tex]
plug in the known values:
[tex] \frac{8.6}{\sin(90)} = \frac{BC}{\sin(45.75)} [/tex]
Since sin(90)=1, the denominator of the first fraction disappears. Finally, we can solve for BC:
[tex] BC = 8.6 \cdot \sin(45.75) \approx 8.4338\ldots [/tex]
Which gives 8.43 when rounded as required.
