Answer:
To figure out the largest angle of the triangle above, we are going to use cosine rule below
[tex]\begin{gathered} a^2=b^2+c^2-2\times b\times c\cos A \\ \text{where,} \\ a=34.2,b=21.3,c=20.2 \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} a^2=b^2+c^2-2\times b\times c\cos A \\ 34.2^2=21.3^2+20.2^2-2\times21.3\times20.2\cos A^{} \\ 1169.64=453.69+408.04-860.52\cos A \end{gathered}[/tex]
Collect similar terms
[tex]\begin{gathered} 1169.64=453.69+408.04-860.52\cos A \\ 1169.64=861.73-860.52\cos A \\ 860.52\cos A=861.73-1169.64 \\ \frac{860.52\cos A}{860.52}=\frac{-307.91}{860.52} \\ \cos A=-0.3578 \\ A=\cos ^{-1}-0.3578 \\ A=69.03^0(first\text{ quadrant)} \\ F\in al\text{ answr will be} \\ A=180^0-69.03^0 \\ A=110.97^0 \end{gathered}[/tex]
Hence,
The largest angle is = 110.97°
