Geometrico 3gdl
- 2. 𝜃1 = 𝑡𝑎𝑛−1 (
𝑑𝑦
𝑑𝑥
)
∅ = 𝑡𝑎𝑛−1(
dz
√dy2 + dx2
)
𝛽 = 90° − ∅
𝜔 = √ 𝑥2 + 𝑦2 + 𝑧2
𝑟 = √1002 + 𝜔2 − 2(100)(𝜔)cos( 𝛽)
𝛾 = 𝑠𝑒𝑛−1 (
𝜔𝑠𝑒𝑛(𝛽)
𝑟
)
𝜀 = 𝑐𝑜𝑠−1 (
−𝑟2 + 1752 + 1402
2(175)(140)
)
𝜃3 = 270° − 𝜀
𝜇 = 𝑠𝑒𝑛−1 (
140𝑠𝑒𝑛(𝜀)
𝑟
)
𝜃2 = 𝜇 + 𝛾