题型概述：方位角与相对速度Question type: bearings & relative velocity

1方位角（bearing）是什么What a bearing is

方位角从正北开始、顺时针量到运动方向，写成三位数（如 068°）。它和数学课的角度（从正东逆时针）不同，所以分量公式也反过来：A bearing is measured from North, clockwise, to the direction of travel and is written with three digits (e.g. 068°). It differs from the maths angle (anticlockwise from East), so the component formulas swap:

为什么东是 sin：β = 0（正北）时东分量应为 0，而 sin 0 = 0。用这个"特殊值检验"就不会记混。Why East uses sin: at β = 0 (due North) the East component must be 0, and sin 0 = 0. This special-value check stops you mixing them up.

2相对速度：站到别人的船上看Relative velocity: stand on the other ship

"A 看到 B 的速度"= B 的速度减去 A 的速度。直觉：站在 A 的甲板上，A 自己不动，整个世界都以 −v_A 在动，所以 B 看上去的速度是 v_B + (−v_A)。"The velocity of B as seen from A" is B's velocity minus A's. Intuition: on A's deck A is still and the whole world moves at −v_A, so B appears to move at v_B + (−v_A).

第二种写法最常用："真实速度 = 观察者速度 + 看到的相对速度"。题目说"relative to the ferry, C sails on bearing 160°"，指的是 v_C − v_F 的方向是 160°。The second form is the workhorse: "true velocity = observer's velocity + observed relative velocity". "Relative to the ferry, C sails on 160°" means the direction of v_C − v_F is 160°.

3两种解法：推荐"坐标分量法"Two methods — prefer components

两种方法评分标准都给满分。几何法（画三角形 + 正弦/余弦定理）画图快，但 arcsin 有两个解，容易掉进钝角陷阱；坐标法（全部化成东/北分量再加减）步骤机械、不会出方向错误，更适合考场。Mark schemes accept both. The geometric method (triangle + sine/cosine rules) is quick to sketch, but arcsin has two roots and invites the obtuse-angle trap. The component method (everything in East/North components) is mechanical and direction-safe, which is better under exam pressure.