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Using cos Law, we can easily find TU.

\(TU = \sqrt{24^2+12^2-2*24*12*cos(38)}\)

TU ≈ 16.3127497964741677

Note: in cos(38), the 38 is in degrees, not radians.

This is a Law of Cosines problem.

Law of Cosines:

\(c^2 = a^2 + b^2 - 2ab\cos C \\ \\ c = TU, a=24, b=12, C=38^{\circ} \\ \text{ (a and b are interchangeable, just pick one and stick with it.)} \\ TU^2 =24^2 +12^2-2(24)(12)\cos38 \\ TU^2 = 576 + 144 - 576\cos38 \\ TU^2 = 720 - 576\cos38 \\ TU = \sqrt{720 - 576\cos38} \\ TU \approx 16.3 \text{ mm}\)

Help

Let \(\angle{TUV} = \varphi\)

\(\begin{array}{|rcll|} \hline \tan(\varphi) &=& \frac{24\cdot \sin(38^{\circ})}{12-24\cdot \cos(38^{\circ})} \\ \tan(\varphi) &=& \frac{24\cdot 0.61566147533}{12-24\cdot 0.78801075361} \\ \tan(\varphi) &=& \frac{14.7758754078}{-6.91225808656} \quad & | \quad \text{II. Quadrant}\\ \tan(\varphi) &=& -2.13763363908 \\ \varphi &=& -64.9294774677^{\circ} + 180^{\circ} \\ \varphi &=& 115.070522532^{\circ} \\ \hline \end{array} \)

TU = ?

\(\begin{array}{|rcll|} \hline \frac{ \sin(38^{\circ}) } {TU} &=& \frac{ \sin(\varphi) } {24} \\ \frac{ \sin(38^{\circ}) } {TU} &=& \frac{ \sin(115.070522532^{\circ}) } {24} \\ TU &=& 24\cdot \frac{\sin(38^{\circ})}{\sin(115.070522532^{\circ}) } \\ TU &=& 24\cdot \frac{0.61566147533}{0.90578692079 } \\ TU &=& 16.3127497965 \\ \hline \end{array} \)

\(TU \approx 16.3\ mm\)