求解三角聯立方程式...

2007-12-13 12:51 am

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!!

回答 (3)

2007-12-14 11:53 am

✔ 最佳答案

This is indeed a geometry question, as implied by the previous answer.

If we analyse each of the two equations, we have three points, P1, P2 and P3. The coordinates of the points are P1(x1,y1), P2(x2,y2), P3(x3,y3).

The equation
a= arctan((Y3-y)/(X3-x))-arctan((Y2-y)/(X2-x))
means that the angle subtended by the points P2 and P3 at the point P(x,y) is a.
This also implies that points P2 and P3 are on the circumference of a circle such that the point P is also on the circle where the subtended angle is always a, and that subtended at the centre is 2a.
Thus if we draw a circle of suitable radius r1, P(x,y) has to lie somewhere on the circle C1. The radius can be found by the formula
r1=sqrt((x3-x2)^2+(y3-y2)^2)/(2sin(b))

Similarly, if we draw another circle C2 of radius r2 that passes through P1, P2, P should be somewhere on the circumference of the circle. The radius r2 can similarly be found by
r2=sqrt((x2-x1)^2+(y2-y1)^2)/(2sin(a))

If we construct the two circles C1 and C2, the intersection points are P2 (known) and P, and hence P can be solved.

Note that for each pair of points (P1-P2, and P2-P3), there are two possible circles, the centre of which is on each side of the line. Thus there are four possible solutions to the problem.

See
http://abaeterno.reallyfreehosting.com/
for an illustration.

👍 0 👎 0

doraemonpaul

2007-12-18 10:31 am

參考: My Maths knowledge

👍 0 👎 0

用戶19935

2007-12-13 10:09 pm

let simplify:

a= arctan[A]-arctan[B]
b= arctan[B]-arctan[C]

tan[a]= (A-B)/(1+AB)
tan[b]= (B-C)/(1+BC)

how ever, this is cannot be solved, coz
(A-B)/(1+AB)=(x3 (y - y2) + x (y2 - y3) + x2 (-y + y3))/(x^2 + x2 x3 -
x (x2 + x3) + (y - y2) (y - y3))
the 分母 has x and y product.

the way out is, set (x/y) be your new unknown, and

(1+AB) tan[a]=A-B is an quadratic function of (x/y)

Too complicated.

may be, post your original question.
Is your 三角聯立方程式 from some Geometry question?

👍 0 👎 0

收錄日期: 2021-04-13 14:42:09
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071212000051KK01859

檢視 Wayback Machine 備份