1. **Problem statement:** We need to find the measure of angle $S$ in triangle $STR$ where side $TR=4.7$ mi, side $SR=5.1$ mi, and angle $T=77^\circ$. 2. **Formula used:** We will use the Law of Cosines to find angle $S$. The Law of Cosines states: $$\cos(S) = \frac{a^2 + c^2 - b^2}{2ac}$$ where $a$, $b$, and $c$ are the sides of the triangle, and $S$ is the angle opposite side $b$. 3. **Assign sides:** Let - $a = SR = 5.1$ mi - $b = TR = 4.7$ mi (side opposite angle $S$) - $c$ is the side opposite angle $T$, which is side $ST$ (unknown length) 4. **Find side $ST$ using Law of Cosines at angle $T$:** $$c^2 = a^2 + b^2 - 2ab \cos(T)$$ $$c^2 = 5.1^2 + 4.7^2 - 2 \times 5.1 \times 4.7 \times \cos(77^\circ)$$ Calculate: $$5.1^2 = 26.01$$ $$4.7^2 = 22.09$$ $$\cos(77^\circ) \approx 0.224951$$ $$c^2 = 26.01 + 22.09 - 2 \times 5.1 \times 4.7 \times 0.224951$$ $$c^2 = 48.1 - 10.77 = 37.33$$ $$c = \sqrt{37.33} \approx 6.11$$ 5. **Now find angle $S$ using Law of Cosines:** $$\cos(S) = \frac{a^2 + c^2 - b^2}{2ac} = \frac{5.1^2 + 6.11^2 - 4.7^2}{2 \times 5.1 \times 6.11}$$ Calculate numerator: $$26.01 + 37.33 - 22.09 = 41.25$$ Calculate denominator: $$2 \times 5.1 \times 6.11 = 62.32$$ $$\cos(S) = \frac{41.25}{62.32} \approx 0.6618$$ 6. **Find angle $S$:** $$S = \cos^{-1}(0.6618) \approx 48.5^\circ$$ 7. **Final answer:** The measure of angle $S$ is approximately $49^\circ$ to the nearest degree.