Solve the triangle.
\[
\mathrm{a}=8.208 \text { in } \mathrm{c}=6.659 \text { in } \mathrm{B}=78.02^{\circ}
\]
What is the length of side $b$ ?
in
(Round to the nearest thousandth as needed.)
What is the measure of angle $A$ ?
(Round to the nearest hundredth as needed.)
What is the measure of angle $\mathrm{C}$ ?
(Round to the nearest hundredth as needed.)

Step 1 ：We are given two sides and an included angle of a triangle. We can use the Law of Cosines to find the length of side $b$. The Law of Cosines states that $b^2 = a^2 + c^2 - 2ac \cos B$.

Step 2 ：Substitute the given values into the formula: $b^2 = (8.208)^2 + (6.659)^2 - 2*(8.208)*(6.659) \cos (78.02)$

Step 3 ：Solve the equation to find the value of $b$, which is approximately 9.435 inches.

Step 4 ：After finding $b$, we can use the Law of Sines to find the other two angles. The Law of Sines states that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.

Step 5 ：Use the Law of Sines to find angle $A$: $A = \sin^{-1}(\frac{a \sin B}{b}) = \sin^{-1}(\frac{8.208 \sin 78.02}{9.435})$, which is approximately 58.32 degrees.

Step 6 ：Use the Law of Sines to find angle $C$: $C = \sin^{-1}(\frac{c \sin B}{b}) = \sin^{-1}(\frac{6.659 \sin 78.02}{9.435})$, which is approximately 43.66 degrees.

Step 7 ：Final Answer: The length of side $b$ is approximately \(\boxed{9.435}\) in. The measure of angle $A$ is approximately \(\boxed{58.32}\) degrees. The measure of angle $C$ is approximately \(\boxed{43.66}\) degrees.