Question - Triangle Geometry Problem

Given triangle ABC with angle A = 32° and sides AC = 8 and BC = 5.5, we will use the Law of Sines to find side AB.

According to the Law of Sines:

\[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \]

Let AB = c, then:

\[ \frac{c}{\sin(32°)} = \frac{8}{\sin(B)} \] \end{p>

To find angle B, we can use the sine rule again:

\[ \frac{5.5}{\sin(B)} = \frac{8}{\sin(32°)} \end{p>

Rearranging gives:

\[ \sin(B) = \frac{5.5 \cdot \sin(32°)}{8} \end{p>

Calculating sin(B) and then angle B, we can find angle C = 180° - A - B.

Finally, using angle C, apply the Law of Sines again to find side c:

\[ \frac{c}{\sin(C)} = \frac{5.5}{\sin(B)} \end{p>