To find the measures requested, we can use the properties of angle bisectors and the given information.
First, let's find \(m \angle SAV\). Since \(AV\) is an angle bisector, it divides \(m \angle BAC\) into two equal angles. Therefore, \(m \angle SAV = \frac{1}{2} m \angle BAC = \frac{1}{2} \cdot 64^\circ = 32^\circ\).
Next, let's find \(m \angle SAU\). Since \(AU\) is also an angle bisector, it divides \(m \angle BAC\) into two equal angles. Therefore, \(m \angle SAU = \frac{1}{2} m \angle BAC = \frac{1}{2} \cdot 64^\circ = 32^\circ\).
Now, let's find \(m \angle SBV\). Since \(BV\) is an angle bisector, it divides \(m \angle ABC\) into two equal angles. Therefore, \(m \angle SBV = \frac{1}{2} m \angle ABC = \frac{1}{2} \cdot 22^\circ = 11^\circ\).
Finally, let's find \(SV\). We can use the Angle Bisector Theorem, which states that the ratio of the lengths of the segments formed by an angle bisector is equal to the ratio of the lengths of the opposite sides. In this case, we have:
\(\frac{SV}{TV} = \frac{CV}{CU}\)
Substituting the given values, we get:
\(\frac{SV}{18} = \frac{27}{CU}\)
Solving for \(SV\), we find:
\(SV = \frac{27}{CU} \cdot 18\)
Since \(m \angle TCU = 22^\circ\), we can use the Law of Sines to find \(CU\):
\(\frac{CU}{\sin(22^\circ)} = \frac{TV}{\sin(180^\circ - 32^\circ - 22^\circ)}\)
Simplifying, we get:
\(\frac{CU}{\sin(22^\circ)} = \frac{18}{\sin(126^\circ)}\)
Solving for \(CU\), we find:
\(CU \approx \frac{\sin(22^\circ) \cdot 18}{\sin(126^\circ)}\)
Substituting this value back into the equation for \(SV\), we can find \(SV\).
Therefore, the final answers are:
\(m \angle SAU = 32^\circ\)
\(m \angle SBV = 11^\circ\)
\(SV \approx \frac{\sin(22^\circ) \cdot 18}{\sin(126^\circ)}\) (approximately)