To solve this question we will use the Sine Law:
[tex]\frac{\sin \measuredangle G}{g}=\frac{\sin \measuredangle F}{f}=\frac{\sin \measuredangle E}{e}\text{.}[/tex]
Substituting g=970in, f=930in, and ∡F=121° we get:
[tex]\frac{\sin\measuredangle G}{970in}=\frac{\sin121^{\circ}}{930in}\text{.}[/tex]
Multiplying the above equation by 970in we get:
[tex]\begin{gathered} \frac{\sin\measuredangle G}{970in}\times970in=\frac{\sin121^{\circ}}{930in}\times970in, \\ \sin \measuredangle G=\frac{970}{930}\sin 121^{\circ}\text{.} \end{gathered}[/tex]
Therefore:
[tex]\begin{gathered} \measuredangle G=\sin ^{-1}(\frac{970}{930}\sin 121^{\circ}) \\ \measuredangle G\approx63^{\circ}. \end{gathered}[/tex]
Answer:
[tex]63^{\circ}.[/tex]
