Solve the triangle.
\[
a=3, b=9, c=8
\]
\(\boxed{A = \sin^{-1}(\frac{3*\sin(\cos^{-1}(\frac{10}{27}))}{8}), B = 180 - A - C, C = \cos^{-1}(\frac{10}{27})}\)
Step 1 :\(\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}\)
Step 2 :Substitute the given values into the equation: \(\cos(C) = \frac{3^2 + 9^2 - 8^2}{2*3*9} = \frac{10}{27}\)
Step 3 :Calculate the angle C: \(C = \cos^{-1}(\frac{10}{27})\)
Step 4 :Use the Law of Sines to find another angle: \(\sin(A) = \frac{a*\sin(C)}{c}\)
Step 5 :Substitute the given values into the equation: \(\sin(A) = \frac{3*\sin(\cos^{-1}(\frac{10}{27}))}{8}\)
Step 6 :Calculate the angle A: \(A = \sin^{-1}(\frac{3*\sin(\cos^{-1}(\frac{10}{27}))}{8})\)
Step 7 :Find the third angle by subtracting the other two angles from 180 degrees: \(B = 180 - A - C\)
Step 8 :So, the solution to the triangle is \(A = \sin^{-1}(\frac{3*\sin(\cos^{-1}(\frac{10}{27}))}{8})\), \(B = 180 - A - C\), and \(C = \cos^{-1}(\frac{10}{27})\)
Step 9 :\(\boxed{A = \sin^{-1}(\frac{3*\sin(\cos^{-1}(\frac{10}{27}))}{8}), B = 180 - A - C, C = \cos^{-1}(\frac{10}{27})}\)