An angle bisector of a triangle divides the opposite side of the triangle into segments 6 cm and 8 cm long. A second side of the triangle measures 7 cm long. Find the longest and shortest possible lengths of the third side of the triangle. Round answers to the nearest tenth. (1 point)
A 8.3 cm and 5.8 cm
B 5.3 cm and 9.3 cm
C 4.2 cm and 7.6 cm
D 7.4 cm and 9.3 cm
Let $AC$ be the third side of the triangle, with $A$ opposite to the 6 cm segment and $C$ opposite to the 8 cm segment. By the angle bisector theorem, we have $\frac{AC}{CA'}=\frac{6}{8}=\frac{3}{4}$, where $CA'$ is the length of the angle bisector.
Let $BC=7$ cm and $BA'=x$. Then by the angle bisector theorem again, we have $\frac{BA'}{A'C}=\frac{7}{AC}=\frac{7}{x}$. Combining the two ratios gives us $\frac{7}{x}=\frac{3}{4}$. Cross multiplying gives $4\cdot 7=3\cdot x$, so $x=\frac{28}{3}$.
Now, applying the triangle inequality, we have $AC<AB+BC$, or $AC<\frac{28}{3}+7=\frac{49}{3} \approx 16.3$. Similarly, we have $AC>BC-AB$, or $AC>7-\frac{28}{3}=\frac{7}{3} \approx 2.3$.
Therefore, the longest and shortest possible lengths of $AC$ are $\boxed{\textbf{(A)}\ 8.3\ \text{cm and}\ 5.8\ \text{cm}}$.