Solution
We need to convert Angle C to only degrees.
Since
[tex]\begin{gathered} 1^0=60^{\prime} \\ \Rightarrow x^{}=10^{\prime} \\ \\ \Rightarrow1^0\times10^{\prime}=60^{\prime}\times x \\ \\ \Rightarrow x=\frac{10}{60}=\frac{1}{6}^0 \end{gathered}[/tex][tex]\Rightarrow C=91^010^{\prime}=(91+\frac{1}{6})^0=\frac{547}{6}^0[/tex][tex]\begin{gathered} \text{ The area of a triangle A} \\ A=\frac{1}{2}a\cdot b\sin C \\ \\ \Rightarrow A=\frac{1}{2}\cdot24\cdot5\cdot\sin (\frac{547}{6})\approx59.99 \end{gathered}[/tex]
