Refer to the diagram below.
Let x = length of line segment RS.
Let Q = ∠RQS.
Let S = ∠RSQ
According to the Law of Sines,
[tex] \frac{sinS}{2.4} = \frac{sin80^{o}}{3.1} \\\\sinS = (\frac{2.4}{3.1}) sin 80^{o}= 0.7624 \\\\ S = sin^{-1}0.7624 = 49.68^{o}[/tex]
Because the sum of angles in a triangle is 180°, therefore
∠Q = 180 -(80 + 49.68) = 50.32°
Apply the Law of Sines again to obtain
[tex] \frac{x}{sinQ} = \frac{3.1}{sin80^{o}} \\\\ x = ( \frac{sin50.32^{o}}{sin80^{o}})3.1 = 2.423[/tex]
Answer: The length of RS = 2.4 (nearest tenth)
