20. In $\triangle \mathrm{ABC}$, where $\angle \mathrm{A}=24^{\circ}, a=60$, and $b=90$, which of the following solves the triangle where $c$ is the largest side? \[ \begin{array}{l} \text { A } \angle \mathrm{A}=24^{\circ}, \angle \mathrm{B}=52^{\circ}, \angle \mathrm{C}=104^{\circ}, a=60, b=90, c=119 \\ \text { B } \angle \mathrm{A}=24^{\circ}, \angle \mathrm{B}=65^{\circ}, \angle \mathrm{C}=91^{\circ}, a=60, b=90, c=110 \\ \text { C } \angle \mathrm{A}=24^{\circ}, \angle \mathrm{B}=142^{\circ}, \angle \mathrm{C}=14^{\circ}, a=60, b=90, c=36 \\ \text { D } \angle \mathrm{A}=24^{\circ}, \angle \mathrm{B}=38^{\circ}, \angle \mathrm{C}=118^{\circ}, a=60, b=90, c=130 \end{array} \]

Step 1 ：\(\angle A = 24^\circ, a = 60, b = 90\)

Step 2 ：Use the Law of Sines: \(\frac{a}{\sin{A}} = \frac{b}{\sin{B}}\)

Step 3 ：Plug in the given values: \(\frac{60}{\sin{24^\circ}} = \frac{90}{\sin{B}}\)

Step 4 ：Solve for \(\sin{B}\): \(\sin{B} = \frac{90 \times \sin{24^\circ}}{60}\)

Step 5 ：Calculate \(\sin{B}\): \(\sin{B} \approx 0.6428\)

Step 6 ：Find \(\angle B\): \(\angle B \approx 40^\circ\)

Step 7 ：Calculate \(\angle C\): \(\angle C = 180^\circ - \angle A - \angle B \approx 116^\circ\)

Step 8 ：Use the Law of Cosines to find \(c\): \(c^2 = a^2 + b^2 - 2ab\cos{C}\)

Step 9 ：Plug in the given values: \(c^2 = 60^2 + 90^2 - 2(60)(90)\cos{116^\circ}\)

Step 10 ：Calculate \(c^2\): \(c^2 \approx 16900\)

Step 11 ：Find \(c\): \(c \approx 130\)

Step 12 ：The triangle is: \(\angle A = 24^\circ, \angle B = 40^\circ, \angle C = 116^\circ, a = 60, b = 90, c = 130\)

Step 13 ：\(\boxed{\text{D}}\)