Beschrijving
Problem
Given is a right triangle
△
A
B
C
{\displaystyle \triangle ABC}
with side length
a
0
=
|
A
B
|
{\displaystyle a_{0}=|AB|}
and the
∠
β
{\displaystyle \angle \beta }
in vertex B. These two elements determine the triangle.
What is the general formula to determine the side length
a
1
{\displaystyle a_{1}}
of any square
◻
D
E
F
G
{\displaystyle \square DEFG}
within the triangle provided that at least three vertices of the square touch the triangle?
Solution
Since at least three vertices of the square touch the triangle there must be a point
D
{\displaystyle D}
where the square touches the line
A
B
¯
{\displaystyle {\overline {AB}}}
. Similarily the vertices
E
{\displaystyle E}
and
G
{\displaystyle G}
touch the triangle sides
A
C
¯
{\displaystyle {\overline {AC}}}
and
B
C
¯
{\displaystyle {\overline {BC}}}
respectively.
So for any inscribed
◻
D
E
F
G
{\displaystyle \square DEFG}
there is a right
△
A
D
G
{\displaystyle \triangle ADG}
with two of the square's vertices and the vertex
A
{\displaystyle A}
of the base triangle
△
A
B
C
{\displaystyle \triangle ABC}
.
The triangle
△
A
D
G
{\displaystyle \triangle ADG}
has an angle
∠
δ
{\displaystyle \angle \delta }
in the vertex
D
{\displaystyle D}
. Or put the other way: To any angle
δ
{\displaystyle \delta }
with
0
<=
δ
<=
π
2
{\displaystyle 0<=\delta <={\frac {\pi }{2}}}
there is a specific square
◻
D
E
F
G
{\displaystyle \square DEFG}
that satisfies the condition of being "three-touchy".
Within the
△
A
D
G
{\displaystyle \triangle ADG}
the angle in vertex
G
{\displaystyle G}
has the size
π
−
π
2
−
δ
=
π
2
−
δ
{\displaystyle \pi -{\frac {\pi }{2}}-\delta ={\frac {\pi }{2}}-\delta }
.
Concerning the vertices
E
{\displaystyle E}
and
F
{\displaystyle F}
the following has to be considered:
As long as
0
<=
δ
<=
β
{\displaystyle 0<=\delta <=\beta }
then the vertex
E
{\displaystyle E}
touches the side
B
C
¯
{\displaystyle {\overline {BC}}}
of
△
A
B
C
{\displaystyle \triangle ABC}
(see Figure 1).
Once
δ
>
β
{\displaystyle \delta >\beta }
then vertex
E
{\displaystyle E}
dissolves from the side and it is vertex
F
{\displaystyle F}
that touches
B
C
¯
{\displaystyle {\overline {BC}}}
of
△
A
B
C
{\displaystyle \triangle ABC}
.
Solution for 0 <= δ <= β
General solution
Drawing a perpendicular from triangle side
A
B
¯
{\displaystyle {\overline {AB}}}
to vertex
E
{\displaystyle E}
gives two further right triangles
△
D
H
E
{\displaystyle \triangle DHE}
and
△
B
E
H
{\displaystyle \triangle BEH}
.
Within the
△
D
H
E
{\displaystyle \triangle DHE}
the angle in vertex
D
{\displaystyle D}
has the size
π
−
π
2
−
δ
=
π
2
−
δ
{\displaystyle \pi -{\frac {\pi }{2}}-\delta ={\frac {\pi }{2}}-\delta }
. So the angle in
E
{\displaystyle E}
is math>\delta</math>.
Considering the above delivers the following equations:
(1)
a
1
=
|
D
G
|
=
|
F
G
|
=
|
E
F
|
=
|
D
E
|
{\displaystyle \quad a_{1}=|DG|=|FG|=|EF|=|DE|}
(2)
a
0
=
|
A
B
|
=
|
A
D
|
+
|
D
H
|
+
|
B
H
|
{\displaystyle \quad a_{0}=|AB|=|AD|+|DH|+|BH|}
(3)
|
A
D
|
|
D
G
|
=
c
o
s
(
δ
)
{\displaystyle \quad {\frac {|AD|}{|DG|}}=cos{(\delta )}}
(4)
|
E
H
|
|
D
E
|
=
c
o
s
(
δ
)
{\displaystyle \quad {\frac {|EH|}{|DE|}}=cos{(\delta )}}
(5)
|
D
H
|
|
D
E
|
=
s
i
n
(
δ
)
{\displaystyle \quad {\frac {|DH|}{|DE|}}=sin{(\delta )}}
(6)
|
E
H
|
|
B
H
|
=
t
a
n
(
β
)
{\displaystyle \quad {\frac {|EH|}{|BH|}}=tan{(\beta )}}
Starting with equation (2) we have:
a
0
=
|
A
D
|
+
|
D
H
|
+
|
B
H
|
{\displaystyle \quad a_{0}=|AD|+|DH|+|BH|}
⇔
a
0
=
|
D
G
|
⋅
c
o
s
(
δ
)
+
|
D
H
|
+
|
B
H
|
{\displaystyle \quad \Leftrightarrow a_{0}=|DG|\cdot cos{(\delta )}+|DH|+|BH|\quad }
, applying equation (3)
⇔
a
0
=
a
1
⋅
c
o
s
(
δ
)
+
|
D
H
|
+
|
B
H
|
{\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+|DH|+|BH|\quad }
, applying equation (1)
⇔
a
0
=
a
1
⋅
c
o
s
(
δ
)
+
|
D
E
|
⋅
s
i
n
(
δ
)
+
|
B
H
|
{\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+|DE|\cdot sin{(\delta )}+|BH|\quad }
, applying equation (5)
⇔
a
0
=
a
1
⋅
c
o
s
(
δ
)
+
a
1
⋅
s
i
n
(
δ
)
+
|
B
H
|
{\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+a_{1}\cdot sin{(\delta )}+|BH|\quad }
, applying equation (5)
⇔
a
0
=
a
1
⋅
c
o
s
(
δ
)
+
a
1
⋅
s
i
n
(
δ
)
+
|
E
H
|
t
a
n
(
β
)
{\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+a_{1}\cdot sin{(\delta )}+{\frac {|EH|}{tan{(\beta )}}}\quad }
, applying equation (6)
⇔
a
0
=
a
1
⋅
c
o
s
(
δ
)
+
a
1
⋅
s
i
n
(
δ
)
+
|
D
E
|
⋅
c
o
s
(
δ
)
t
a
n
(
β
)
{\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+a_{1}\cdot sin{(\delta )}+{\frac {|DE|\cdot cos{(\delta )}}{tan{(\beta )}}}\quad }
, applying equation (4)
⇔
a
0
=
a
1
⋅
c
o
s
(
δ
)
+
a
1
⋅
s
i
n
(
δ
)
+
a
1
⋅
c
o
s
(
δ
)
t
a
n
(
β
)
{\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot cos{(\delta )}+a_{1}\cdot sin{(\delta )}+{\frac {a_{1}\cdot cos{(\delta )}}{tan{(\beta )}}}\quad }
, applying equation (1)
⇔
a
0
=
a
1
⋅
(
c
o
s
(
δ
)
+
s
i
n
(
δ
)
+
c
o
s
(
δ
)
t
a
n
(
β
)
)
{\displaystyle \quad \Leftrightarrow a_{0}=a_{1}\cdot \left(cos{(\delta )}+sin{(\delta )}+{\frac {cos{(\delta )}}{tan{(\beta )}}}\right)\quad }
, rearranging
⇔
a
1
=
a
0
c
o
s
(
δ
)
+
s
i
n
(
δ
)
+
c
o
s
(
δ
)
t
a
n
(
β
)
{\displaystyle \quad \Leftrightarrow a_{1}={\frac {a_{0}}{cos{(\delta )}+sin{(\delta )}+{\frac {cos{(\delta )}}{tan{(\beta )}}}}}\quad }
, rearranging
⇔
a
1
=
a
0
s
i
n
(
δ
)
+
c
o
s
(
δ
)
⋅
(
1
+
1
t
a
n
(
β
)
)
{\displaystyle \quad \Leftrightarrow a_{1}={\frac {a_{0}}{sin{(\delta )}+cos{(\delta )}\cdot \left(1+{\frac {1}{tan{(\beta )}}}\right)}}\quad }
, rearranging
Special solutions
If
δ
=
0
{\displaystyle \delta =0\quad }
then
a
1
=
a
0
(
1
+
1
tan
β
)
{\displaystyle a_{1}={\frac {a_{0}}{\left(1+{\frac {1}{\tan \beta }}\right)}}}
If
δ
=
β
{\displaystyle \delta =\beta \quad }
then
a
1
=
a
0
sin
β
+
cos
β
⋅
(
1
+
1
tan
β
)
=
a
0
sin
β
+
(
cos
β
+
cos
β
tan
β
)
=
a
0
sin
β
+
cos
β
+
cos
β
tan
β
{\displaystyle a_{1}={\frac {a_{0}}{\sin \beta +\cos \beta \cdot \left(1+{\frac {1}{\tan \beta }}\right)}}={\frac {a_{0}}{\sin \beta +\left(\cos \beta +{\frac {\cos \beta }{\tan \beta }}\right)}}={\frac {a_{0}}{\sin \beta +\cos \beta +{\frac {\cos \beta }{\tan \beta }}}}}
If
β
=
π
4
{\displaystyle \beta ={\frac {\pi }{4}}}
(i.e. the triangle
△
A
B
C
{\displaystyle \triangle ABC}
is a right isosceles triangle) then:
a
1
=
a
0
sin
δ
+
cos
δ
⋅
(
1
+
1
1
)
=
a
0
sin
δ
+
2
⋅
cos
δ
{\displaystyle a_{1}={\frac {a_{0}}{\sin \delta +\cos \delta \cdot \left(1+{\frac {1}{1}}\right)}}={\frac {a_{0}}{\sin \delta +2\cdot \cos \delta }}}
If
δ
=
0
{\displaystyle \delta =0}
and
β
=
π
4
{\displaystyle \beta ={\frac {\pi }{4}}\quad }
then
a
1
=
a
0
2
{\displaystyle a_{1}={\frac {a_{0}}{2}}}
If
δ
=
β
=
π
4
{\displaystyle \delta =\beta ={\frac {\pi }{4}}\quad }
then
a
1
=
a
0
1
2
+
2
⋅
1
2
=
a
0
3
⋅
1
2
=
2
⋅
a
0
3
{\displaystyle a_{1}={\frac {a_{0}}{{\frac {1}{\sqrt {2}}}+2\cdot {\frac {1}{\sqrt {2}}}}}={\frac {a_{0}}{3\cdot {\frac {1}{\sqrt {2}}}}}={\frac {{\sqrt {2}}\cdot a_{0}}{3}}}
Extremal points
For a given
β
{\displaystyle \beta }
the function
a
1
=
a
0
sin
δ
+
cos
δ
⋅
(
1
+
1
tan
β
)
{\displaystyle a_{1}={\frac {a_{0}}{\sin \delta +\cos \delta \cdot \left(1+{\frac {1}{\tan \beta }}\right)}}}
can be transformed to
a
1
=
a
0
sin
δ
+
k
⋅
cos
δ
{\displaystyle a_{1}={\frac {a_{0}}{\sin \delta +k\cdot \cos \delta }}\quad }
with
k
=
1
+
1
tan
β
{\displaystyle \quad k=1+{\frac {1}{\tan \beta }}\quad }
.
The derivative is:
d
a
1
d
δ
=
a
0
⋅
−
1
(
sin
δ
+
k
⋅
cos
δ
)
2
⋅
(
cos
δ
−
k
⋅
sin
δ
)
=
a
0
⋅
k
⋅
sin
δ
−
cos
δ
(
sin
δ
+
k
⋅
cos
δ
)
2
{\displaystyle \quad {\frac {da_{1}}{d\delta }}=a_{0}\cdot {\frac {-1}{\left(\sin \delta +k\cdot \cos \delta \right)^{2}}}\cdot \left(\cos \delta -k\cdot \sin \delta \right)=a_{0}\cdot {\frac {k\cdot \sin \delta -\cos \delta }{\left(\sin \delta +k\cdot \cos \delta \right)^{2}}}}
For
δ
=
0
{\displaystyle \delta =0\quad }
the derivative is:
d
a
1
d
δ
=
0
=
a
0
⋅
k
⋅
0
−
1
(
0
+
k
⋅
1
)
2
=
−
a
0
k
2
<
0
{\displaystyle \quad {\frac {da_{1}}{d\delta _{=0}}}=a_{0}\cdot {\frac {k\cdot 0-1}{\left(0+k\cdot 1\right)^{2}}}={\frac {-a_{0}}{k^{2}}}<0}
For
δ
=
β
{\displaystyle \delta =\beta \quad }
the derivative is:
d
a
1
d
δ
=
β
=
a
0
⋅
k
⋅
sin
β
−
cos
β
(
sin
β
+
k
⋅
cos
β
)
2
=
a
0
⋅
(
1
+
1
tan
β
)
⋅
sin
β
−
cos
β
(
sin
β
+
k
⋅
cos
β
)
2
=
a
0
⋅
(
sin
β
+
sin
β
tan
β
)
−
cos
β
(
sin
β
+
k
⋅
cos
β
)
2
=
a
0
⋅
sin
β
+
cos
β
−
cos
β
(
sin
β
+
k
⋅
cos
β
)
2
>
0
{\displaystyle \quad {\frac {da_{1}}{d\delta _{=\beta }}}=a_{0}\cdot {\frac {k\cdot \sin \beta -\cos \beta }{\left(\sin \beta +k\cdot \cos \beta \right)^{2}}}=a_{0}\cdot {\frac {\left(1+{\frac {1}{\tan \beta }}\right)\cdot \sin \beta -\cos \beta }{\left(\sin \beta +k\cdot \cos \beta \right)^{2}}}=a_{0}\cdot {\frac {\left(\sin \beta +{\frac {\sin \beta }{\tan \beta }}\right)-\cos \beta }{\left(\sin \beta +k\cdot \cos \beta \right)^{2}}}=a_{0}\cdot {\frac {\sin \beta +\cos \beta -\cos \beta }{\left(\sin \beta +k\cdot \cos \beta \right)^{2}}}>0}
So in the interval
0
<=
δ
<=
β
{\displaystyle 0<=\delta <=\beta }
the derivative changes from negative to positive. This means there is a
δ
=
ϕ
{\displaystyle \delta =\phi }
with
d
a
1
d
δ
=
ϕ
=
0
{\displaystyle {\frac {da_{1}}{d\delta _{=\phi }}}=0}
To find
ϕ
{\displaystyle \phi }
, we put
d
a
1
d
δ
=
ϕ
=
0
⇒
a
0
⋅
k
⋅
sin
ϕ
−
cos
ϕ
(
sin
ϕ
+
k
⋅
cos
ϕ
)
2
=
0
⇒
k
⋅
sin
ϕ
−
cos
ϕ
=
0
⇒
k
⋅
sin
ϕ
=
cos
ϕ
⇒
k
⋅
sin
ϕ
cos
ϕ
=
1
{\displaystyle {\frac {da_{1}}{d\delta _{=\phi }}}=0\quad \Rightarrow \quad a_{0}\cdot {\frac {k\cdot \sin \phi -\cos \phi }{\left(\sin \phi +k\cdot \cos \phi \right)^{2}}}=0\quad \Rightarrow \quad k\cdot \sin \phi -\cos \phi =0\quad \Rightarrow \quad k\cdot \sin \phi =\cos \phi \quad \Rightarrow \quad k\cdot {\frac {\sin \phi }{\cos \phi }}=1}
⇒
(
1
+
1
tan
β
)
⋅
tan
ϕ
=
1
⇒
(
tan
β
+
1
tan
β
)
⋅
tan
ϕ
=
1
⇒
tan
ϕ
=
tan
β
tan
β
+
1
{\displaystyle \quad \Rightarrow \quad \left(1+{\frac {1}{\tan \beta }}\right)\cdot \tan \phi =1\quad \Rightarrow \quad \left({\frac {\tan \beta +1}{\tan \beta }}\right)\cdot \tan \phi =1\quad \Rightarrow \quad \tan \phi ={\frac {\tan \beta }{\tan \beta +1}}}
So, in the intervall
0
<=
δ
<=
β
{\displaystyle \quad 0<=\delta <=\beta \quad }
the smallest square with at least three vertices touching the surrounding right triangle occurs when
tan
δ
=
tan
β
tan
β
+
1
{\displaystyle \tan \delta ={\frac {\tan \beta }{\tan \beta +1}}}
In the case of a right isosceles triangle with
β
=
π
4
{\displaystyle \beta ={\frac {\pi }{4}}}
the smallest square occurs when
tan
δ
=
1
1
+
1
=
0.5
⇒
δ
≈
0.464
π
{\displaystyle \quad \tan \delta ={\frac {1}{1+1}}=0.5\quad \Rightarrow \quad \delta \approx 0.464\pi }
Solution for β <= δ <= 90°
General solution
Drawing a perpendicular from triangle side
A
C
¯
{\displaystyle {\overline {AC}}}
to vertex
F
{\displaystyle F}
gives two further right triangles
△
G
F
J
{\displaystyle \triangle GFJ}
and
△
C
J
F
{\displaystyle \triangle CJF}
.
Within the
△
G
F
J
{\displaystyle \triangle GFJ}
the angle in vertex
G
{\displaystyle G}
has the size
π
−
π
2
−
(
π
2
−
δ
)
=
δ
{\displaystyle \pi -{\frac {\pi }{2}}-({\frac {\pi }{2}}-\delta )=\delta }
.
The triangle
△
C
J
F
{\displaystyle \triangle CJF}
is similar to the triangle
△
A
B
C
{\displaystyle \triangle ABC}
Considering the above delivers the following equations:
(1)
a
1
=
|
D
G
|
=
|
F
G
|
{\displaystyle \quad a_{1}=|DG|=|FG|}
(2)
|
A
C
|
=
|
A
G
|
+
|
G
J
|
+
|
C
J
|
=
a
0
⋅
tan
β
{\displaystyle \quad |AC|=|AG|+|GJ|+|CJ|=a_{0}\cdot \tan \beta }
(3)
|
A
G
|
|
D
G
|
=
|
A
G
|
a
1
=
sin
δ
⇒
|
A
G
|
=
a
1
⋅
sin
δ
{\displaystyle \quad {\frac {|AG|}{|DG|}}={\frac {|AG|}{a_{1}}}=\sin \delta \quad \Rightarrow |AG|=a_{1}\cdot \sin \delta }
(4)
|
G
J
|
|
F
G
|
=
|
G
J
|
a
1
=
cos
δ
⇒
|
G
J
|
=
a
1
⋅
cos
δ
{\displaystyle \quad {\frac {|GJ|}{|FG|}}={\frac {|GJ|}{a_{1}}}=\cos \delta \quad \Rightarrow |GJ|=a_{1}\cdot \cos \delta }
(5)
|
C
J
|
|
F
J
|
=
|
A
C
|
|
A
B
|
=
tan
β
⇒
|
C
J
|
=
|
F
J
|
⋅
tan
β
=
a
1
⋅
sin
δ
⋅
tan
β
{\displaystyle \quad {\frac {|CJ|}{|FJ|}}={\frac {|AC|}{|AB|}}=\tan \beta \quad \Rightarrow |CJ|=|FJ|\cdot \tan \beta =a_{1}\cdot \sin \delta \cdot \tan \beta }
Applying equations (3), (4) and (5) to equation (2) leads to
a
1
⋅
sin
δ
+
a
1
⋅
cos
δ
+
a
1
⋅
sin
δ
⋅
tan
β
=
a
0
⋅
tan
β
{\displaystyle a_{1}\cdot \sin \delta +a_{1}\cdot \cos \delta +a_{1}\cdot \sin \delta \cdot \tan \beta =a_{0}\cdot \tan \beta }
⇔
a
1
⋅
(
sin
δ
+
cos
δ
+
sin
δ
⋅
tan
β
)
=
a
0
⋅
tan
β
{\displaystyle \quad \Leftrightarrow a_{1}\cdot (\sin \delta +\cos \delta +\sin \delta \cdot \tan \beta )=a_{0}\cdot \tan \beta }
⇔
a
1
⋅
(
cos
δ
+
sin
δ
⋅
(
1
+
tan
β
)
)
=
a
0
⋅
tan
β
{\displaystyle \quad \Leftrightarrow a_{1}\cdot (\cos \delta +\sin \delta \cdot (1+\tan \beta ))=a_{0}\cdot \tan \beta }
⇔
a
1
=
a
0
⋅
tan
β
cos
δ
+
sin
δ
⋅
(
1
+
tan
β
)
{\displaystyle \quad \Leftrightarrow a_{1}={\frac {a_{0}\cdot \tan \beta }{\cos \delta +\sin \delta \cdot (1+\tan \beta )}}}
Special solutions
If
δ
=
π
2
{\displaystyle \delta ={\frac {\pi }{2}}\quad }
then
a
1
=
a
0
⋅
tan
β
0
+
1
⋅
(
1
+
tan
β
)
=
a
0
⋅
tan
β
1
+
tan
β
=
a
0
1
tan
β
+
1
=
a
0
1
+
1
tan
β
{\displaystyle a_{1}={\frac {a_{0}\cdot \tan \beta }{0+1\cdot (1+\tan \beta )}}={\frac {a_{0}\cdot \tan \beta }{1+\tan \beta }}={\frac {a_{0}}{{\frac {1}{\tan \beta }}+1}}={\frac {a_{0}}{1+{\frac {1}{\tan \beta }}}}}
If
δ
=
β
{\displaystyle \delta =\beta \quad }
then
a
1
=
a
0
⋅
tan
β
cos
β
+
sin
β
⋅
(
1
+
tan
β
)
=
a
0
cos
β
tan
β
+
sin
β
tan
β
⋅
(
1
+
tan
β
)
=
a
0
cos
2
β
sin
β
+
cos
β
⋅
(
1
+
tan
β
)
{\displaystyle \quad a_{1}={\frac {a_{0}\cdot \tan \beta }{\cos \beta +\sin \beta \cdot (1+\tan \beta )}}={\frac {a_{0}}{{\frac {\cos \beta }{\tan \beta }}+{\frac {\sin \beta }{\tan \beta }}\cdot (1+\tan \beta )}}={\frac {a_{0}}{{\frac {\cos ^{2}\beta }{\sin \beta }}+\cos \beta \cdot (1+\tan \beta )}}}
If
β
=
π
4
{\displaystyle \beta ={\frac {\pi }{4}}\quad }
(i.e. the triangle
△
A
B
C
{\displaystyle \triangle ABC}
is a right isosceles triangle) then
a
1
=
a
0
⋅
1
cos
δ
+
sin
δ
⋅
(
1
+
1
)
=
a
0
cos
δ
+
2
⋅
sin
δ
{\displaystyle \quad a_{1}={\frac {a_{0}\cdot 1}{\cos \delta +\sin \delta \cdot (1+1)}}={\frac {a_{0}}{\cos \delta +2\cdot \sin \delta }}}
If
δ
=
π
2
{\displaystyle \delta ={\frac {\pi }{2}}}
and
β
=
π
4
{\displaystyle \beta ={\frac {\pi }{4}}\quad }
then
⇔
a
1
=
a
0
⋅
1
0
+
1
⋅
(
1
+
1
)
=
a
0
2
{\displaystyle \quad \Leftrightarrow a_{1}={\frac {a_{0}\cdot 1}{0+1\cdot (1+1)}}={\frac {a_{0}}{2}}}
If
δ
=
β
=
π
4
{\displaystyle \delta =\beta ={\frac {\pi }{4}}\quad }
then
a
1
=
a
0
⋅
1
1
2
+
1
2
⋅
(
1
+
1
)
=
2
3
⋅
a
0
{\displaystyle a_{1}={\frac {a_{0}\cdot 1}{{\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {2}}}\cdot (1+1)}}={\frac {\sqrt {2}}{3}}\cdot a_{0}}
Position of the vertices
Putting
A
{\displaystyle A}
on
0
+
0
i
{\displaystyle 0+0i}
, will put the other vertices in both figures to:
P
o
s
(
B
)
=
a
0
+
0
i
{\displaystyle Pos(B)=a_{0}+0i}
P
o
s
(
C
)
=
0
+
(
a
0
⋅
tan
β
)
i
{\displaystyle Pos(C)=0+(a_{0}\cdot \tan \beta )i}
P
o
s
(
D
)
=
(
a
1
⋅
cos
δ
)
+
0
i
{\displaystyle Pos(D)=(a_{1}\cdot \cos \delta )+0i}
P
o
s
(
E
)
=
(
|
A
D
|
+
|
A
H
|
)
+
(
|
E
H
|
)
i
=
(
a
1
⋅
cos
δ
+
a
1
⋅
sin
δ
)
+
(
|
E
H
|
)
i
=
(
a
1
⋅
cos
δ
+
a
1
⋅
sin
δ
)
+
(
a
1
⋅
cos
δ
)
i
{\displaystyle Pos(E)=(|AD|+|AH|)+(|EH|)i=(a_{1}\cdot \cos \delta +a_{1}\cdot \sin \delta )+(|EH|)i=(a_{1}\cdot \cos \delta +a_{1}\cdot \sin \delta )+(a_{1}\cdot \cos \delta )i}
P
o
s
(
F
)
=
(
|
F
J
|
)
+
(
|
A
G
|
+
|
G
J
|
)
i
=
(
a
1
⋅
sin
δ
)
+
(
|
A
G
|
+
|
G
J
|
)
i
=
(
a
1
⋅
sin
δ
)
+
(
a
1
⋅
sin
δ
+
a
1
⋅
cos
δ
)
i
{\displaystyle Pos(F)=(|FJ|)+(|AG|+|GJ|)i=(a_{1}\cdot \sin \delta )+(|AG|+|GJ|)i=(a_{1}\cdot \sin \delta )+(a_{1}\cdot \sin \delta +a_{1}\cdot \cos \delta )i}
P
o
s
(
G
)
=
0
+
(
a
1
⋅
sin
δ
)
i
{\displaystyle Pos(G)=0+(a_{1}\cdot \sin \delta )i}
Licentie
Ik, de auteursrechthebbende van dit werk, maak het hierbij onder de volgende licentie beschikbaar:
De gebruiker mag:
Delen – het werk kopiëren, verspreiden en doorgeven
Remixen – afgeleide werken maken
Onder de volgende voorwaarden:
naamsvermelding – U moet op een gepaste manier aan naamsvermelding doen, een link naar de licentie geven, en aangeven of er wijzigingen in het werk zijn aangebracht. U mag dit op elke redelijke manier doen, maar niet zodanig dat de indruk wordt gewekt dat de licentiegever instemt met uw werk of uw gebruik van zijn werk.
Gelijk delen – Als u het werk heeft geremixt, veranderd, of erop heeft voortgebouwd, moet u het gewijzigde materiaal verspreiden onder dezelfde licentie als het oorspronkelijke werk, of een daarmee compatibele licentie . https://creativecommons.org/licenses/by-sa/4.0 CC BY-SA 4.0 Creative Commons Attribution-Share Alike 4.0 true true Nederlands Beschrijf in één regel wat dit bestand voorstelt