简单数学工具备用

/// 计算3个点之间的角度
///
/// - Parameters:
///   - p1: 一端
///   - p2: 另一端
///   - p0: 中间点
/// - Returns: 角度 单位:°(度)
public func computeAngle(_ p1: CGPoint,_ p2: CGPoint,_ p0: CGPoint) -> CGFloat{
    let dx1 = p1.x - p0.x;
    let dy1 = p1.y - p0.y;
    let dx2 = p2.x - p0.x;
    let dy2 = p2.y - p0.y;
    let cos_angle = (dx1*dx2 + dy1*dy2)/sqrt((dx1*dx1 + dy1*dy1)*(dx2*dx2 + dy2*dy2) + 1e-10)
    return acos(cos_angle) * 180.0 / CGFloat.pi
    
}

/// 一般式 直线方程  Ax + By + C = 0
public typealias LinearEquation = (CGFloat,CGFloat,CGFloat)

/// 计算直线方程 一般式 Ax + By + C = 0
///
/// - Parameters:
///   - p1: 第一个点
///   - p2: 第二个点
/// - Returns: 一般式方程
public func computeLinearEquation(_ p1: CGPoint,_ p2: CGPoint) -> LinearEquation {
    return (p2.y-p1.y,p1.x-p2.x,p1.y*p2.x-p1.x*p2.y)
}


/// 两线的交点
///
/// - Parameters:
///   - line1: 第一条一般式直线方程
///   - line2: 第二条一般式直线方程
/// - Returns: 交点
public func compute2lineIntersection(_ line1: LinearEquation, _ line2: LinearEquation) -> CGPoint?{
    //斜率不存在
    if line1.1 == 0 && line2.1 == 0 {
        return nil
    }
    //斜率相等
    if line1.0 / line1.1 == line2.0 / line2.1 {
        return nil
    }
    
    //有交点
    let x = (line1.1*line2.2 - line2.1*line1.2)/(line1.0*line2.1 - line2.0*line1.1)
    let y = (line2.2*line1.0 - line1.2*line2.0)/(line2.0*line1.1 - line1.0*line2.1)
    return CGPoint(x: x, y: y)    
}