Question

The distance from City A to City B is 430 kilometers, that from City B to City C 835 kilometers, and that from City C to City A 910 kilometers. What are the angles in the triangle with these three cities as vertices?

Answer 1

Question

$$AB = 430$$ $$BC = 835$$ $$CA = 910$$

Solving:

$$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos \angle B \Rightarrow$$ $$\Rightarrow \cos \angle B = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} \Rightarrow \angle B = \arccos \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}.$$ $$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos \angle A \Rightarrow$$ $$\Rightarrow \cos \angle A = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC} \Rightarrow \angle A = \arccos \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC}.$$ $$\angle C = 180{}^\circ - \angle A - \angle B$$

So, we have:

$$\angle A = \arccos \frac{430^2 + 910^2 - 835^2}{2 \cdot 430 \cdot 910} = \arccos \frac{315775}{782600} \approx 66{}^\circ.$$ $$\angle B = \arccos \frac{430^2 + 835^2 - 910^2}{2 \cdot 430 \cdot 835} = \arccos \frac{54025}{718100} \approx 86{}^\circ.$$ $$\angle C = 180{}^\circ - 66{}^\circ - 86{}^\circ = 28{}^\circ.$$

Answer: $\angle A = 66{}^\circ; \angle B = 86{}^\circ; \angle C = 28{}^\circ$.