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?
\[
\begin{array}{l}
\text { A } \angle \mathrm{A}=24^{\circ}, \angle \mathrm{B}=52^{\circ}, \angle \mathrm{C}=104^{\circ}, a=60, b=90, c=119 \\
\text { B } \angle \mathrm{A}=24^{\circ}, \angle \mathrm{B}=65^{\circ}, \angle \mathrm{C}=91^{\circ}, a=60, b=90, c=110 \\
\text { C } \angle \mathrm{A}=24^{\circ}, \angle \mathrm{B}=142^{\circ}, \angle \mathrm{C}=14^{\circ}, a=60, b=90, c=36 \\
\text { D } \angle \mathrm{A}=24^{\circ}, \angle \mathrm{B}=38^{\circ}, \angle \mathrm{C}=118^{\circ}, a=60, b=90, c=130
\end{array}
\]
\(\boxed{\text{D}}\)
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 :\(\boxed{\text{D}}\)