1. **State the problem:** We have two triangles, EBR and BRO, sharing side BR = 24.58 km. We want to find the unknown side EB in triangle EBR using given angles and sides. 2. **Identify known values:** - In triangle EBR: angle B = 64.6°, side BR = 24.58 km, side EB = ? - In triangle BRO: sides BR = 24.58 km, RO = 51.76 km, BO = 57.3 km, angle O = 25.4° - EO = 98 km (connecting E and O) 3. **Use Law of Cosines in triangle BRO to find angle B:** $$\cos B = \frac{BR^2 + BO^2 - RO^2}{2 \times BR \times BO} = \frac{24.58^2 + 57.3^2 - 51.76^2}{2 \times 24.58 \times 57.3}$$ Calculate numerator: $$24.58^2 = 604.0964, \quad 57.3^2 = 3283.29, \quad 51.76^2 = 2679.4976$$ $$604.0964 + 3283.29 - 2679.4976 = 1207.889$$ Calculate denominator: $$2 \times 24.58 \times 57.3 = 2817.468$$ So, $$\cos B = \frac{1207.889}{2817.468} \approx 0.4285$$ 4. **Find angle B:** $$B = \cos^{-1}(0.4285) \approx 64.6^\circ$$ This matches the given angle B in triangle EBR, confirming consistency. 5. **Use Law of Sines in triangle EBR to find EB:** Law of Sines formula: $$\frac{EB}{\sin 64.6^\circ} = \frac{BR}{\sin E}$$ We need angle E. Since EO = 98 km connects E and O, and angles at E and O are unknown, we approximate angle E by considering triangle EBR and BRO combined or use given data. 6. **Assuming angle E is supplementary to angle O (25.4°) in triangle ERO, approximate angle E:** $$E = 180^\circ - 25.4^\circ = 154.6^\circ$$ 7. **Calculate EB:** $$EB = \frac{BR \times \sin 64.6^\circ}{\sin 154.6^\circ} = \frac{24.58 \times \sin 64.6^\circ}{\sin 154.6^\circ}$$ Calculate sines: $$\sin 64.6^\circ \approx 0.9004, \quad \sin 154.6^\circ = \sin (180^\circ - 154.6^\circ) = \sin 25.4^\circ \approx 0.4290$$ So, $$EB = \frac{24.58 \times 0.9004}{0.4290} = \frac{22.13}{0.4290} \approx 51.58 \text{ km}$$ **Final answer:** $$\boxed{EB \approx 51.58 \text{ km}}$$