Solution:
Given:
[tex]\begin{gathered} First\text{ arc:} \\ \theta=315^0 \\ r=11 \\ \\ \\ \\ \\ Second\text{ arc:} \\ \theta=210^0 \\ r=14 \end{gathered}[/tex]
To find the length of an arc, the formula below is used:
[tex]l=\frac{\theta}{360}\times2\pi r[/tex]
Hence, for the first arc;
[tex]\begin{gathered} l=\frac{315}{360}\times2\times\pi\times11 \\ l=\frac{6930\pi}{360} \\ l=60.476 \\ \\ To\text{ the nearest tenth,} \\ l\approx60.5 \end{gathered}[/tex]
Therefore, the length of the first arc to the nearest tenth is 60.5 units
Hence, for the second arc;
[tex]\begin{gathered} l=\frac{210}{360}\times2\times\pi\times14 \\ l=\frac{5880\pi}{360} \\ l=51.313 \\ \\ To\text{ the nearest tenth,} \\ l\approx51.3 \end{gathered}[/tex]
Therefore, the length of the second arc to the nearest tenth is 51.3 units
