测量距离问题（不可达两点、隔河测距）

A 在河对岸，B、C 在同岸，量得 BC = 100 m

测得 ∠ABC = π/6，∠ACB = π/3

∠A = π − π/6 − π/3 = π/2

AB = BC·sin∠ACB/sin∠A = 100×(√3/2)/1 = 50√3 (m)

答：AB = 50√3 m ≈ 86.6 m

A、B 是池塘两端，中间不能穿越，但两端都能到达

取岸上一点 C，量得 CA = 50 m，CB = 80 m，∠ACB = π/3

余弦定理：AB² = CA² + CB² − 2×CA×CB×cos∠ACB

AB² = 2500 + 6400 − 2×50×80×(1/2) = 4900

答：AB = 70 m

A、B 都在河对岸，本岸取 C、D，量得 CD = 100 m

测得 ∠ACD = 2π/3，∠CDA = π/6；∠BCD = π/6，∠CDB = π/3

△ACD：∠CAD = π − 2π/3 − π/6 = π/6

AC = CD·sin∠CDA/sin∠CAD = 100×sin(π/6)/sin(π/6) = 100 (m)

△BCD：∠CBD = π − π/6 − π/3 = π/2，BC = 100×sin(π/3)/sin(π/2) = 50√3 (m)

∠ACB = ∠ACD − ∠BCD = 2π/3 − π/6 = π/2

AB² = AC² + BC² = 100² + (50√3)² = 10000 + 7500 = 17500

AB = 50√7 ≈ 132.3 (m)

复算：△ABD 中 AD = 100√3、BD = 50、∠ADB = π/3 − π/6 = π/6

AB² = 30000 + 2500 − 2×100√3×50×cos(π/6) = 17500，两条路一致