求解三角聯立方程式...

Question

求解三角聯立方程式...
1. a = arctan[(Y3-y)/(X3-x)]-arctan[(Y2-y)/(X2-x)]
2. b = arctan[(Y2-y)/(X2-x)]-arctan[(Y1-y)/(X1-x)]

上式中a,b,X1,X2,X3,Y3,Y2,Y1為已知,求解x,y? ..謝謝!!

Answer 1

設A(x1, y1), B(x2, y2), C(x3, y3)


1. 滿足1. 條件的(x,y)為某圓的一部份, 此圓作法如下以BC線段為一邊,作中垂線,
取高為tan(a/2)*BC/2(請注意是哪一邊!),得一三角形,作此三角形的外接圓


2. 同上步驟再作一圓


3. 以上兩圓的交點即為所求
註:先求公共弦之中點坐標,及公共弦的長度,再以向量方法求交點

Answer 2

A = tan(a)
= [(Y3-y)/(X3-x) - (Y2-y)/(X2-x)]/{1+[(Y3-y)/(X3-x)][(Y2-y)/(X2-x)]
B = tan(b)
= [(Y2-y)/(X2-x) - (Y1-y)/(X1-x)]/{1+[(Y2-y)/(X2-x)][(Y1-y)/(X1-x)]
然後化成兩個(聯立)二次方程式.