Construct the trapezoid ABCD ,where \(AB || CD\)

\(AB = 12\) and \(CD = 22\)

angle \(C=65^{\circ}\) and angle \(D = 40^{\circ}\)

Draw \(AE || BC\) where \(E\) is on \(CD\)

So now \(ABCE\) is a parallelogram, and \(CE = 12\)

which makes \(ED = 10\)

Now look at triangle \(AED,\) by corresponding angles

angle \(AED = 65°\), angle \(D = 40\) leaving angle \(DAE = 75°\)

by sine law:

\(\frac{AD}{\sin65} = \frac{10}{\sin75}\)

\(AD = \frac{10\sin65}{\sin75} = 9.38\)

by sine law:

\(AE \sin40 = {10}{\sin75}\)

\(AE = 6.65\)

but \(BC = AE\),

So the side adjacent to the \(65°\) angle is \(6.65\), the side adjacent to the \(40°\) angle is \(9.38\)