三角測量(29)俯角
Author: 藍邦偉  老師

已知塔高，由塔上H點觀測湖上兩船P，Q，得俯角α與β，測兩船之距離。

三角測量(29)俯角

u = Vector((x(Corner[1.0]), 0.0), (x(Corner[3.0]), 0.0))
v = Vector((0.0, y(Corner[1.0])), (0.0, y(Corner[3.0])))
A = (8.445950413223137, 0.6560330578512381)
B = (10.682314049586772, 0.765123966942147)
C = (8.464132231404957, -2.3257851239669427)
D = Corner(2.0)
多邊形1, a, b, d, c = Polygon(A, B, D, C)
E = (8.445950413223137, 3.783305785123964)
F = (10.70049586776859, 3.492396694214873)
多邊形2, a_1, b_1, f, e = Polygon(A, B, F, E)
G = (8.464132231404955, 5.8196694214876)
H_3 = Corner(3.0)
多邊形3, h, g, e_1, f_1 = Polygon(H_3, G, E, F)
O = (8.336859504132228, 0.5651239669421472)
P = (9.282314049586773, -0.39851239669421606)
P_1: P + (0.8, 0.0)
P_2: 0.5 * P + 0.5 * P_1 + (0.0, -(0.3))
k = CircumcircleArc(P, P_2, P_1)
Q = (9.009586776859502, -1.7076033057851245)
Q_1: Q + (0.8, 0.0)
Q_2: 0.5 * Q + 0.5 * Q_1 + (0.0, -(0.3))
p = CircumcircleArc(Q, Q_2, Q_1)
A_1: 0.5 * P + 0.5 * P_1
A_2: A_1 + (0.0, 0.4)
多邊形4, i, a_2, a_3, I = Polygon(A_2, A_1, 3.0)
B_1: 0.5 * Q + 0.5 * Q_1
B_2: B_1 + (0.0, 0.4)
多邊形5, j, b_2, b_3, J = Polygon(B_1, B_2, 3.0)
H = (1.2641322314049588, 1.7469421487603285)
K: H + (0.0, -(3.7))
l = Segment(H, K)
m = Segment(K, Q_2)
n = Segment(K, P_2)
q = Segment(H, P_2)
r = Segment(H, Q_2)
s = Line(H, n)
t = Line(H, m)
L: x = x(Corner[2.0])
H_1 = Intersect(L, s)
H_2 = Intersect(L, t)
c_1 = Segment(H, H_1)
d_1 = Segment(H, H_2)
α = Angle(H_1, H, P_2)
β = Angle(H_2, H, Q_2)
M = (8.445950413223137, 4.092396694214873)
N = (9.082314049586774, 4.328760330578509)
R = (9.755041322314044, 3.8378512396694187)
S = (10.664132231404954, 4.056033057851237)
多邊形6, m_1, e_2, f_2, s_1, r_1, n_1 = Polygon(M, E, F, S, R, N)