Aleksandra Ciepielowska

Sprawozdanie nr 4

Zadanie 1. Obliczyć Azymut A₁₂ oraz odległość d₁₂ mając dane

1. X₁=4,08 x₂=28,21
   y₁=32,46 y₂=56,18

2. X₁= 42,81 x₂=4,95
   y₁= 26,56 y₂=112,17

3. X₁= 86,21 x₂=4,26
   y₁= 48,52 y₂=4,84

4. X₁= 12,56 x₂=76,18
   y₁= 32,84 y₂=4,72

Rozwiązanie

∆x₁₂= x₂-x₁

∆y₁₂=y₂-y₁

$$d_{12} = \sqrt{{(x_{12})}^{2} + {(y_{12})}^{2}}$$

$$\varphi = arctg(\frac{y_{12}}{{x}_{12}})$$

1. ∆x₁₂=28,21-4,08=24,13m

∆y₁₂=56,18-32,46=23,72m

d₁₂=$\sqrt{\left( 24,13 \right)^{2} + \left( 23,72 \right)^{2}}$=33,84m

$$\varphi = \text{arctg}\left( \frac{23,72}{24,13} \right) = 49,4545$$

∆x₁₂>0 i ∆y₁₂>0 => ćwiartka I
A₁₂=ϕ

A₁₂=49,4545^g

1. ∆x₁₂=-37,86m

∆y₁₂=85,61m

d₁₂=$\sqrt{\left( - 37,86 \right)^{2} + \left( 85,61 \right)^{2}}$=93,61m

$$\varphi = arctg\left( \frac{85,61}{- 37,86} \right) = - 73,4924$$

∆x₁₂<0 i ∆y₁₂>0 => ćwiartka II

A₁₂=ϕ+200^g

A₁₂=-73,4924+200^g=126,5076^g

1. ∆x₁₂=-81,95m

∆y₁₂=-43,68m

d₁₂=$\sqrt{\left( - 81,95 \right)^{2} + \left( - 43,68 \right)^{2}}$=92,86m

$$\varphi = arctg\left( \frac{- 81,95}{- 43,68} \right) = 31,1755$$

∆x₁₂<0 i ∆y₁₂<0 => ćwiartka III
A₁₂=ϕ+200^g

A₁₂=231,1755^g

1. ∆x₁₂=63,62m

∆y₁₂=-28,12m

d₁₂=$\sqrt{\left( 63,62 \right)^{2} + \left( - 28,12 \right)^{2}}$=92,86m

$$\varphi = arctg\left( \frac{- 28,12}{+ 63,62} \right) = - 26,4949$$

∆x₁₂>0 i ∆y₁₂<0 => ćwiartka IV
A₁₂=ϕ+400^g

A₁₂=373,5051^g

Zadanie 2. Obliczyć współrzędne punktu 3 leżącego na prostej 12 mając dane:
x₁=4,26 x₂=36,42

y₁=4,14 y₂=118,24

d₁₃=4,82m

∆x₁₂=32,16

∆y₁₂=114,10

$\varphi = \text{arctg}\left( \frac{114,10}{32,16} \right) = 82,5101$

∆x₁₂>0 i ∆y₁₂>0 => ćwiartka I

A₁₂=ϕ

A₁₂=82,5101

A₁₂= A₁₃

A₁₃=82,5101

∆x₁₃=x₃-x₁ => x₃=∆x₁₃+x₁

X₃=x₁+d₁₃*cosA₁₃

X₃=4,26+4,82*cos(82,5101)=5,57

∆y₁₃=y₃-y₁ => y₃=∆y₁₃+y₁

y₃=y₁+d_13*sinA₁₃

y₃=4,14+4,82*sin(82,5101)=8,78

Zadanie 3. Oblicz współrzędne punktu 3 mając dane:

x₁=4,09 x₂=126,18

y₁=4,82 y₂=196,24

α=28^g20^c17^cc

β=35^g41^c81^cc

∆x₁₂=122,09

∆y₁₂=191,42

$$\varphi = arctg\left( \frac{191,42}{122,09} \right) = 63,8553$$

A₁₂=ϕ

A₁₂=63,8553^g

A₁₃=A₁₂-α=35,6536^g

∆x₂₁=-122,09

∆y₂₁=-191,42

$$\varphi = arctg\left( \frac{- 91,42}{- 122,09} \right) = 63,8553$$

A₂₁=ϕ+200^g

A₂₁=263,8553^g

A₂₃= A₂₁+β

A₂₃= 299,2734

d₁₂ =227,04

$\frac{d_{13}}{\text{sinβ}} = \frac{d_{12}}{sin(\alpha + \beta)}$

$d_{13} = \frac{{sin\beta*d}_{12}}{sin(\alpha + \beta)} = 142,54$

x₁₃ = x₃^I − x₁ = > x₃^I = x₁ + x₁₃

x₃^I = x₁ + d₁₃ * cosA₁₃

x₃^I = 4, 09 + 142, 54 * cos(35,6536) = 124, 85

y₁₃ = y₃^I − y₁ = > y₃^I = y₁ + y₁₃

y₃^I = y₁ + d₁₃ * sinA₁₃

y₃^I = 4, 82 + 142, 54 * sin(35, 6536)=80, 54

$$d_{23} = \frac{{sin\alpha*d}_{12}}{sin(\alpha + \beta)} = 115,70$$

x₂₃ = x₃^II − x₂ = > x₃^II = x₂ + x₂₃

x₃^II = x₂ + d₂₃ * cosA₂₃

x₃^II = 126, 18 + 115, 70 * cos(299,2734) = 124, 86

y₂₃ = y₃^II − y₂ = > y₃^II = y₂ + y₂₃

y₃^II = y₂ + d₂₃ * sinA₂₃

y₃^II = 196, 24 + 115, 70 * sin(299,2734) = 80, 55

$$x_{3} = \frac{x_{3}^{I} + x_{3}^{\text{II}}}{2} = 124,86$$

$${\backslash ty}_{3} = \frac{y_{3}^{I} + y_{3}^{\text{II}}}{2} = 80,54$$