English
Let p1, p2, p3 be points with ∠p1 p2 p3 = π/2. Then ∠p2 p3 p1 = arccos( dist(p3,p2) / dist(p1,p3) ).
Русский
Пусть p1, p2, p3 — точки такие, что ∠p1 p2 p3 = π/2. Тогда ∠p2 p3 p1 = arccos( dist(p3,p2) / dist(p1,p3) ).
LaTeX
$$$\angle p_2 p_3 p_1 = \arccos\left( \frac{\operatorname{dist}(p_3,p_2)}{\operatorname{dist}(p_1,p_3)} \right) \quad\text{if } \angle p_1 p_2 p_3 = \frac{\pi}{2}$$$
Lean4
/-- An angle in a right-angled triangle expressed using `arccos`. -/
theorem oangle_right_eq_arccos_of_oangle_eq_pi_div_two {p₁ p₂ p₃ : P} (h : ∡ p₁ p₂ p₃ = ↑(π / 2)) :
∡ p₂ p₃ p₁ = Real.arccos (dist p₃ p₂ / dist p₁ p₃) :=
by
have hs : (∡ p₂ p₃ p₁).sign = 1 := by rw [oangle_rotate_sign, h, Real.Angle.sign_coe_pi_div_two]
rw [oangle_eq_angle_of_sign_eq_one hs,
angle_eq_arccos_of_angle_eq_pi_div_two (angle_eq_pi_div_two_of_oangle_eq_pi_div_two h)]