J u m p t o c o n t e n t
M a i n m e n u
M a i n m e n u
N a v i g a t i o n
● M a i n p a g e
● C o n t e n t s
● C u r r e n t e v e n t s
● R a n d o m a r t i c l e
● A b o u t W i k i p e d i a
● C o n t a c t u s
● D o n a t e
C o n t r i b u t e
● H e l p
● L e a r n t o e d i t
● C o m m u n i t y p o r t a l
● R e c e n t c h a n g e s
● U p l o a d f i l e
S e a r c h
Search
A p p e a r a n c e
● C r e a t e a c c o u n t
● L o g i n
P e r s o n a l t o o l s
● C r e a t e a c c o u n t
● L o g i n
P a g e s f o r l o g g e d o u t e d i t o r s l e a r n m o r e
● C o n t r i b u t i o n s
● T a l k
( T o p )
1
S o l u t i o n m e t h o d o v e r v i e w
2
S o l u t i o n a l g o r i t h m
3
S e e a l s o
4
R e f e r e n c e s
T o g g l e t h e t a b l e o f c o n t e n t s
H a n s e n ' s p r o b l e m
3 l a n g u a g e s
● E s p a ñ o l
● I t a l i a n o
● У к р а ї н с ь к а
E d i t l i n k s
● A r t i c l e
● T a l k
E n g l i s h
● R e a d
● E d i t
● V i e w h i s t o r y
T o o l s
T o o l s
A c t i o n s
● R e a d
● E d i t
● V i e w h i s t o r y
G e n e r a l
● W h a t l i n k s h e r e
● R e l a t e d c h a n g e s
● U p l o a d f i l e
● S p e c i a l p a g e s
● P e r m a n e n t l i n k
● P a g e i n f o r m a t i o n
● C i t e t h i s p a g e
● G e t s h o r t e n e d U R L
● D o w n l o a d Q R c o d e
● W i k i d a t a i t e m
P r i n t / e x p o r t
● D o w n l o a d a s P D F
● P r i n t a b l e v e r s i o n
A p p e a r a n c e
F r o m W i k i p e d i a , t h e f r e e e n c y c l o p e d i a
Fundamental topographical problem
]
In trigonometry , Hansen's problem is a problem in planar surveying , named after the astronomer Peter Andreas Hansen (1795–1874), who worked on the geodetic survey of Denmark. There are two known points A, B , and two unknown points P 1 , P 2 . From P 1 and P 2 an observer measures the angles made by the lines of sight to each of the other three points. The problem is to find the positions of P 1 and P 2 . See figure; the angles measured are (α 1 , β 1 , α 2 , β 2 ) .
Since it involves observations of angles made at unknown points, the problem is an example of resection (as opposed to intersection).
Solution method overview [ edit ]
Define the following angles:
γ
=
∠
P
1
A
P
2
,
δ
=
∠
P
1
B
P
2
,
ϕ
=
∠
P
2
A
B
,
ψ
=
∠
P
1
B
A
.
{\displaystyle {\begin{alignedat}{5}\gamma &=\angle P_{1}AP_{2},&\quad \delta &=\angle P_{1}BP_{2},\\[4pt]\phi &=\angle P_{2}AB,&\quad \psi &=\angle P_{1}BA.\end{alignedat}}}
As a first step we will solve for φ and ψ .
The sum of these two unknown angles is equal to the sum of β 1 and β 2 , yielding the equation
ϕ
+
ψ
=
β
1
+
β
2
.
{\displaystyle \phi +\psi =\beta _{1}+\beta _{2}.}
A second equation can be found more laboriously, as follows. The law of sines yields
A
B
¯
P
2
B
¯
=
sin
α
2
sin
ϕ
,
P
2
B
¯
P
1
P
2
¯
=
sin
β
1
sin
δ
.
{\displaystyle {\frac {\overline {AB}}{\overline {P_{2}B}}}={\frac {\sin \alpha _{2}}{\sin \phi }},\qquad {\frac {\overline {P_{2}B}}{\overline {P_{1}P_{2}}}}={\frac {\sin \beta _{1}}{\sin \delta }}.}
Combining these, we get
A
B
¯
P
1
P
2
¯
=
sin
α
2
sin
β
1
sin
ϕ
sin
δ
.
{\displaystyle {\frac {\overline {AB}}{\overline {P_{1}P_{2}}}}={\frac {\sin \alpha _{2}\sin \beta _{1}}{\sin \phi \sin \delta }}.}
Entirely analogous reasoning on the other side yields
A
B
¯
P
1
P
2
¯
=
sin
α
1
sin
β
2
sin
ψ
sin
γ
.
{\displaystyle {\frac {\overline {AB}}{\overline {P_{1}P_{2}}}}={\frac {\sin \alpha _{1}\sin \beta _{2}}{\sin \psi \sin \gamma }}.}
Setting these two equal gives
sin
ϕ
sin
ψ
=
sin
γ
sin
α
2
sin
β
1
sin
δ
sin
α
1
sin
β
2
=
k
.
{\displaystyle {\frac {\sin \phi }{\sin \psi }}={\frac {\sin \gamma \sin \alpha _{2}\sin \beta _{1}}{\sin \delta \sin \alpha _{1}\sin \beta _{2}}}=k.}
Using a known trigonometric identity this ratio of sines can be expressed as the tangent of an angle difference:
tan
1
2
(
ϕ
−
ψ
)
=
k
−
1
k
+
1
tan
1
2
(
ϕ
+
ψ
)
.
{\displaystyle \tan {\tfrac {1}{2}}(\phi -\psi )={\frac {k-1}{k+1}}\tan {\tfrac {1}{2}}(\phi +\psi ).}
Where
k
=
sin
ϕ
sin
ψ
.
{\displaystyle k={\frac {\sin \phi }{\sin \psi }}.}
This is the second equation we need. Once we solve the two equations for the two unknowns φ, ψ , we can use either of the two expressions above for
A
B
¯
P
1
P
2
¯
{\displaystyle {\tfrac {\overline {AB}}{\overline {P_{1}P_{2}}}}}
to find
P
1
P
2
¯
{\displaystyle {\overline {P_{1}P_{2}}}}
since AB is known. We can then find all the other segments using the law of sines.[1]
Solution algorithm [ edit ]
We are given four angles (α 1 , β 1 , α 2 , β 2 ) and the distance AB . The calculation proceeds as follows:
Calculate
γ
=
π
−
α
1
−
β
1
−
β
2
,
δ
=
π
−
α
2
−
β
1
−
β
2
.
{\displaystyle {\begin{aligned}\gamma &=\pi -\alpha _{1}-\beta _{1}-\beta _{2},\\\delta &=\pi -\alpha _{2}-\beta _{1}-\beta _{2}.\end{aligned}}}
Calculate
k
=
sin
γ
sin
α
2
sin
β
1
sin
δ
sin
α
1
sin
β
2
.
{\displaystyle k={\frac {\sin \gamma \sin \alpha _{2}\sin \beta _{1}}{\sin \delta \sin \alpha _{1}\sin \beta _{2}}}.}
Let
s
=
β
1
+
β
2
,
d
=
2
arctan
(
k
−
1
k
+
1
tan
1
2
s
)
{\displaystyle s=\beta _{1}+\beta _{2},\quad d=2\arctan \left({\frac {k-1}{k+1}}\tan {\tfrac {1}{2}}s\right)}
and then
ϕ
=
s
+
d
2
,
ψ
=
s
−
d
2
.
{\displaystyle \phi ={\frac {s+d}{2}},\quad \psi ={\frac {s-d}{2}}.}
Calculate
P
1
P
2
¯
=
A
B
¯
sin
ϕ
sin
δ
sin
α
2
sin
β
1
{\displaystyle {\overline {P_{1}P_{2}}}={\overline {AB}}\ {\frac {\sin \phi \sin \delta }{\sin \alpha _{2}\sin \beta _{1}}}}
or equivalently
P
1
P
2
¯
=
A
B
¯
sin
ψ
sin
γ
sin
α
1
sin
β
2
.
{\displaystyle {\overline {P_{1}P_{2}}}={\overline {AB}}\ {\frac {\sin \psi \sin \gamma }{\sin \alpha _{1}\sin \beta _{2}}}.}
If one of these fractions has a denominator close to zero, use the other one.
See also [ edit ]
References [ edit ]
R e t r i e v e d f r o m " https://en.wikipedia.org/w/index.php?title=Hansen%27s_problem&oldid=1233841324 "
C a t e g o r i e s :
● T r i g o n o m e t r y
● S u r v e y i n g
● M a t h e m a t i c a l p r o b l e m s
H i d d e n c a t e g o r i e s :
● W e b a r c h i v e t e m p l a t e w a y b a c k l i n k s
● A r t i c l e s w i t h s h o r t d e s c r i p t i o n
● S h o r t d e s c r i p t i o n m a t c h e s W i k i d a t a
● T h i s p a g e w a s l a s t e d i t e d o n 1 1 J u l y 2 0 2 4 , a t 0 5 : 1 2 ( U T C ) .
● T e x t i s a v a i l a b l e u n d e r t h e C r e a t i v e C o m m o n s A t t r i b u t i o n - S h a r e A l i k e L i c e n s e 4 . 0 ;
a d d i t i o n a l t e r m s m a y a p p l y . B y u s i n g t h i s s i t e , y o u a g r e e t o t h e T e r m s o f U s e a n d P r i v a c y P o l i c y . W i k i p e d i a ® i s a r e g i s t e r e d t r a d e m a r k o f t h e W i k i m e d i a F o u n d a t i o n , I n c . , a n o n - p r o f i t o r g a n i z a t i o n .
● P r i v a c y p o l i c y
● A b o u t W i k i p e d i a
● D i s c l a i m e r s
● C o n t a c t W i k i p e d i a
● C o d e o f C o n d u c t
● D e v e l o p e r s
● S t a t i s t i c s
● C o o k i e s t a t e m e n t
● M o b i l e v i e w