the triangle cosine relation in a complex plane The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find a complex number geometricallyShow that for triangle ABC, with complex numbers for the coordinates, that we have the following equationDo sine and cosine of complex numbers have anything to do with right-triangles or circles?Complex plane (Show that triangle is right-angled)Centroid of a Triangle and Cosine LawComplex Number Application in Isosceles Triangle IncentreEquation of a triangle in three dimensional complex plane.Deriving equations regarding triangles in the complex planeRectangular form of $z$ without cosine and sineWhy is the ratio of any two sides of an equilateral triangle on a complex plane equal to a complex cubic root of unity?
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the triangle cosine relation in a complex plane
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find a complex number geometricallyShow that for triangle ABC, with complex numbers for the coordinates, that we have the following equationDo sine and cosine of complex numbers have anything to do with right-triangles or circles?Complex plane (Show that triangle is right-angled)Centroid of a Triangle and Cosine LawComplex Number Application in Isosceles Triangle IncentreEquation of a triangle in three dimensional complex plane.Deriving equations regarding triangles in the complex planeRectangular form of $z$ without cosine and sineWhy is the ratio of any two sides of an equilateral triangle on a complex plane equal to a complex cubic root of unity?
$begingroup$
I have a triangle $ABC$ in a complex plane. The arrangement of vertices is in a counterclockwise direction. The coordinates of $A$,$B$,$C$ are $z_A$,$z_B$,$z_C$ respectively.
It is given that length of side $AB = c, AC = b$ and $angle BAC = alpha$. I need to find the other sides and angles of the triangle. How do i do this with complex numbers?
I know with trigonometry using the sine and cosine rules, all the length and angles can be derived. but how to get it using complex numbers?
complex-numbers analytic-geometry
asked Mar 31 at 5:38
maikmaik
205
$endgroup$
|
show 2 more comments
$begingroup$
I have a triangle $ABC$ in a complex plane. The arrangement of vertices is in a counterclockwise direction. The coordinates of $A$,$B$,$C$ are $z_A$,$z_B$,$z_C$ respectively.
It is given that length of side $AB = c, AC = b$ and $angle BAC = alpha$. I need to find the other sides and angles of the triangle. How do i do this with complex numbers?
I know with trigonometry using the sine and cosine rules, all the length and angles can be derived. but how to get it using complex numbers?
complex-numbers analytic-geometry
asked Mar 31 at 5:38
maikmaik
205
$endgroup$
$begingroup$
my main issue is applying the cosine rule to get the third side. Whats its complex equivalent?
$endgroup$
– maik
Mar 31 at 5:40
$begingroup$
What about $z_B-z_C$?
$endgroup$
– GReyes
Mar 31 at 5:45
$begingroup$
not given, but two sides and the angle in between is known, so the remaining angles and sides can be uniquely determined, at least using trigonometry i know how to find this.
$endgroup$
– maik
Mar 31 at 5:48
$begingroup$
You need $|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)$ and $z_B-z_C=(z_B-z_A)+(z_A-z_C)$. You know that $|z_B-z_A|^2=c^2$ and $|z_C-z_A|^2=b^2$...
$endgroup$
– GReyes
Mar 31 at 5:52
$begingroup$
we also know $angle BAC = alpha$
$endgroup$
– maik
Mar 31 at 5:53
|
show 2 more comments
$begingroup$
I have a triangle $ABC$ in a complex plane. The arrangement of vertices is in a counterclockwise direction. The coordinates of $A$,$B$,$C$ are $z_A$,$z_B$,$z_C$ respectively.
It is given that length of side $AB = c, AC = b$ and $angle BAC = alpha$. I need to find the other sides and angles of the triangle. How do i do this with complex numbers?
I know with trigonometry using the sine and cosine rules, all the length and angles can be derived. but how to get it using complex numbers?
complex-numbers analytic-geometry
asked Mar 31 at 5:38
maikmaik
205
$endgroup$
I have a triangle $ABC$ in a complex plane. The arrangement of vertices is in a counterclockwise direction. The coordinates of $A$,$B$,$C$ are $z_A$,$z_B$,$z_C$ respectively.
It is given that length of side $AB = c, AC = b$ and $angle BAC = alpha$. I need to find the other sides and angles of the triangle. How do i do this with complex numbers?
I know with trigonometry using the sine and cosine rules, all the length and angles can be derived. but how to get it using complex numbers?
complex-numbers analytic-geometry
complex-numbers analytic-geometry
asked Mar 31 at 5:38
maikmaik
205
asked Mar 31 at 5:38
maikmaik
205
asked Mar 31 at 5:38
maikmaik
205
asked Mar 31 at 5:38
maikmaik
205
asked Mar 31 at 5:38
maikmaik
205
205
$begingroup$
my main issue is applying the cosine rule to get the third side. Whats its complex equivalent?
$endgroup$
– maik
Mar 31 at 5:40
$begingroup$
What about $z_B-z_C$?
$endgroup$
– GReyes
Mar 31 at 5:45
$begingroup$
not given, but two sides and the angle in between is known, so the remaining angles and sides can be uniquely determined, at least using trigonometry i know how to find this.
$endgroup$
– maik
Mar 31 at 5:48
$begingroup$
You need $|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)$ and $z_B-z_C=(z_B-z_A)+(z_A-z_C)$. You know that $|z_B-z_A|^2=c^2$ and $|z_C-z_A|^2=b^2$...
$endgroup$
– GReyes
Mar 31 at 5:52
$begingroup$
we also know $angle BAC = alpha$
$endgroup$
– maik
Mar 31 at 5:53
|
show 2 more comments
$begingroup$
my main issue is applying the cosine rule to get the third side. Whats its complex equivalent?
$endgroup$
– maik
Mar 31 at 5:40
$begingroup$
What about $z_B-z_C$?
$endgroup$
– GReyes
Mar 31 at 5:45
$begingroup$
not given, but two sides and the angle in between is known, so the remaining angles and sides can be uniquely determined, at least using trigonometry i know how to find this.
$endgroup$
– maik
Mar 31 at 5:48
$begingroup$
You need $|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)$ and $z_B-z_C=(z_B-z_A)+(z_A-z_C)$. You know that $|z_B-z_A|^2=c^2$ and $|z_C-z_A|^2=b^2$...
$endgroup$
– GReyes
Mar 31 at 5:52
$begingroup$
we also know $angle BAC = alpha$
$endgroup$
– maik
Mar 31 at 5:53
$begingroup$
my main issue is applying the cosine rule to get the third side. Whats its complex equivalent?
$endgroup$
– maik
Mar 31 at 5:40
$begingroup$
my main issue is applying the cosine rule to get the third side. Whats its complex equivalent?
$endgroup$
– maik
Mar 31 at 5:40
$begingroup$
What about $z_B-z_C$?
$endgroup$
– GReyes
Mar 31 at 5:45
$begingroup$
What about $z_B-z_C$?
$endgroup$
– GReyes
Mar 31 at 5:45
$begingroup$
not given, but two sides and the angle in between is known, so the remaining angles and sides can be uniquely determined, at least using trigonometry i know how to find this.
$endgroup$
– maik
Mar 31 at 5:48
$begingroup$
not given, but two sides and the angle in between is known, so the remaining angles and sides can be uniquely determined, at least using trigonometry i know how to find this.
$endgroup$
– maik
Mar 31 at 5:48
$begingroup$
You need $|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)$ and $z_B-z_C=(z_B-z_A)+(z_A-z_C)$. You know that $|z_B-z_A|^2=c^2$ and $|z_C-z_A|^2=b^2$...
$endgroup$
– GReyes
Mar 31 at 5:52
$begingroup$
You need $|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)$ and $z_B-z_C=(z_B-z_A)+(z_A-z_C)$. You know that $|z_B-z_A|^2=c^2$ and $|z_C-z_A|^2=b^2$...
$endgroup$
– GReyes
Mar 31 at 5:52
$begingroup$
we also know $angle BAC = alpha$
$endgroup$
– maik
Mar 31 at 5:53
$begingroup$
we also know $angle BAC = alpha$
$endgroup$
– maik
Mar 31 at 5:53
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Here is the connection with the law of cosines. You can see which quantities correspond to which.
You have $z_B-z_A=ce^ialpha_1$ and $z_C-z_A=be^ialpha_2$, hence $z_A-z_C=be^i(alpha_2+pi)$. Clearly $alpha_2-alpha_1=alpha$. Then
$z_B-z_C=(z_B-z_A)+(z_A-z_C)$ and
$$
a^2=|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)=(z_B-z_A)overline(z_B-z_A)+(z_A-z_C)overline(z_A-z_C)+2Re[(z_B-z_A)overline(z_A-z_C)]=
$$
$$
=c^2+b^2+2bcRe(e^i(alpha_1-alpha_2-pi))=c^2+b^2-2bccosalpha
$$
answered Mar 31 at 6:20
GReyesGReyes
2,48315
$endgroup$
$begingroup$
well i know i can write it using only the real parts. but i want to write it in a purely complex form.
$endgroup$
– maik
Mar 31 at 6:54
$begingroup$
You can replace the term with the real part by $(z_B-z_A)overline(z_A-z_C)+overline(z_B-z_A)(z_A-z_C)$
$endgroup$
– GReyes
Mar 31 at 7:07
$begingroup$
Is this purely complex for you?
$endgroup$
– GReyes
Mar 31 at 7:08
$begingroup$
with modulus , i have to take the square root. I want to avoid sauqre roos and convert to the form with has $e^ialpha$ in it. Then taking the square root will only be dividing the angle in exponential by 2. And i guess it will also look much nicer.
$endgroup$
– maik
Mar 31 at 11:51
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is the connection with the law of cosines. You can see which quantities correspond to which.
You have $z_B-z_A=ce^ialpha_1$ and $z_C-z_A=be^ialpha_2$, hence $z_A-z_C=be^i(alpha_2+pi)$. Clearly $alpha_2-alpha_1=alpha$. Then
$z_B-z_C=(z_B-z_A)+(z_A-z_C)$ and
$$
a^2=|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)=(z_B-z_A)overline(z_B-z_A)+(z_A-z_C)overline(z_A-z_C)+2Re[(z_B-z_A)overline(z_A-z_C)]=
$$
$$
=c^2+b^2+2bcRe(e^i(alpha_1-alpha_2-pi))=c^2+b^2-2bccosalpha
$$
answered Mar 31 at 6:20
GReyesGReyes
2,48315
$endgroup$
$begingroup$
well i know i can write it using only the real parts. but i want to write it in a purely complex form.
$endgroup$
– maik
Mar 31 at 6:54
$begingroup$
You can replace the term with the real part by $(z_B-z_A)overline(z_A-z_C)+overline(z_B-z_A)(z_A-z_C)$
$endgroup$
– GReyes
Mar 31 at 7:07
$begingroup$
Is this purely complex for you?
$endgroup$
– GReyes
Mar 31 at 7:08
$begingroup$
with modulus , i have to take the square root. I want to avoid sauqre roos and convert to the form with has $e^ialpha$ in it. Then taking the square root will only be dividing the angle in exponential by 2. And i guess it will also look much nicer.
$endgroup$
– maik
Mar 31 at 11:51
add a comment |
$begingroup$
Here is the connection with the law of cosines. You can see which quantities correspond to which.
You have $z_B-z_A=ce^ialpha_1$ and $z_C-z_A=be^ialpha_2$, hence $z_A-z_C=be^i(alpha_2+pi)$. Clearly $alpha_2-alpha_1=alpha$. Then
$z_B-z_C=(z_B-z_A)+(z_A-z_C)$ and
$$
a^2=|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)=(z_B-z_A)overline(z_B-z_A)+(z_A-z_C)overline(z_A-z_C)+2Re[(z_B-z_A)overline(z_A-z_C)]=
$$
$$
=c^2+b^2+2bcRe(e^i(alpha_1-alpha_2-pi))=c^2+b^2-2bccosalpha
$$
answered Mar 31 at 6:20
GReyesGReyes
2,48315
$endgroup$
$begingroup$
well i know i can write it using only the real parts. but i want to write it in a purely complex form.
$endgroup$
– maik
Mar 31 at 6:54
$begingroup$
You can replace the term with the real part by $(z_B-z_A)overline(z_A-z_C)+overline(z_B-z_A)(z_A-z_C)$
$endgroup$
– GReyes
Mar 31 at 7:07
$begingroup$
Is this purely complex for you?
$endgroup$
– GReyes
Mar 31 at 7:08
$begingroup$
with modulus , i have to take the square root. I want to avoid sauqre roos and convert to the form with has $e^ialpha$ in it. Then taking the square root will only be dividing the angle in exponential by 2. And i guess it will also look much nicer.
$endgroup$
– maik
Mar 31 at 11:51
add a comment |
$begingroup$
Here is the connection with the law of cosines. You can see which quantities correspond to which.
You have $z_B-z_A=ce^ialpha_1$ and $z_C-z_A=be^ialpha_2$, hence $z_A-z_C=be^i(alpha_2+pi)$. Clearly $alpha_2-alpha_1=alpha$. Then
$z_B-z_C=(z_B-z_A)+(z_A-z_C)$ and
$$
a^2=|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)=(z_B-z_A)overline(z_B-z_A)+(z_A-z_C)overline(z_A-z_C)+2Re[(z_B-z_A)overline(z_A-z_C)]=
$$
$$
=c^2+b^2+2bcRe(e^i(alpha_1-alpha_2-pi))=c^2+b^2-2bccosalpha
$$
answered Mar 31 at 6:20
GReyesGReyes
2,48315
$endgroup$
Here is the connection with the law of cosines. You can see which quantities correspond to which.
You have $z_B-z_A=ce^ialpha_1$ and $z_C-z_A=be^ialpha_2$, hence $z_A-z_C=be^i(alpha_2+pi)$. Clearly $alpha_2-alpha_1=alpha$. Then
$z_B-z_C=(z_B-z_A)+(z_A-z_C)$ and
$$
a^2=|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)=(z_B-z_A)overline(z_B-z_A)+(z_A-z_C)overline(z_A-z_C)+2Re[(z_B-z_A)overline(z_A-z_C)]=
$$
$$
=c^2+b^2+2bcRe(e^i(alpha_1-alpha_2-pi))=c^2+b^2-2bccosalpha
$$
answered Mar 31 at 6:20
GReyesGReyes
2,48315
answered Mar 31 at 6:20
GReyesGReyes
2,48315
answered Mar 31 at 6:20
GReyesGReyes
2,48315
answered Mar 31 at 6:20
GReyesGReyes
2,48315
2,48315
$begingroup$
well i know i can write it using only the real parts. but i want to write it in a purely complex form.
$endgroup$
– maik
Mar 31 at 6:54
$begingroup$
You can replace the term with the real part by $(z_B-z_A)overline(z_A-z_C)+overline(z_B-z_A)(z_A-z_C)$
$endgroup$
– GReyes
Mar 31 at 7:07
$begingroup$
Is this purely complex for you?
$endgroup$
– GReyes
Mar 31 at 7:08
$begingroup$
with modulus , i have to take the square root. I want to avoid sauqre roos and convert to the form with has $e^ialpha$ in it. Then taking the square root will only be dividing the angle in exponential by 2. And i guess it will also look much nicer.
$endgroup$
– maik
Mar 31 at 11:51
add a comment |
$begingroup$
well i know i can write it using only the real parts. but i want to write it in a purely complex form.
$endgroup$
– maik
Mar 31 at 6:54
$begingroup$
You can replace the term with the real part by $(z_B-z_A)overline(z_A-z_C)+overline(z_B-z_A)(z_A-z_C)$
$endgroup$
– GReyes
Mar 31 at 7:07
$begingroup$
Is this purely complex for you?
$endgroup$
– GReyes
Mar 31 at 7:08
$begingroup$
with modulus , i have to take the square root. I want to avoid sauqre roos and convert to the form with has $e^ialpha$ in it. Then taking the square root will only be dividing the angle in exponential by 2. And i guess it will also look much nicer.
$endgroup$
– maik
Mar 31 at 11:51
$begingroup$
well i know i can write it using only the real parts. but i want to write it in a purely complex form.
$endgroup$
– maik
Mar 31 at 6:54
$begingroup$
well i know i can write it using only the real parts. but i want to write it in a purely complex form.
$endgroup$
– maik
Mar 31 at 6:54
$begingroup$
You can replace the term with the real part by $(z_B-z_A)overline(z_A-z_C)+overline(z_B-z_A)(z_A-z_C)$
$endgroup$
– GReyes
Mar 31 at 7:07
$begingroup$
You can replace the term with the real part by $(z_B-z_A)overline(z_A-z_C)+overline(z_B-z_A)(z_A-z_C)$
$endgroup$
– GReyes
Mar 31 at 7:07
$begingroup$
Is this purely complex for you?
$endgroup$
– GReyes
Mar 31 at 7:08
$begingroup$
Is this purely complex for you?
$endgroup$
– GReyes
Mar 31 at 7:08
$begingroup$
with modulus , i have to take the square root. I want to avoid sauqre roos and convert to the form with has $e^ialpha$ in it. Then taking the square root will only be dividing the angle in exponential by 2. And i guess it will also look much nicer.
$endgroup$
– maik
Mar 31 at 11:51
$begingroup$
with modulus , i have to take the square root. I want to avoid sauqre roos and convert to the form with has $e^ialpha$ in it. Then taking the square root will only be dividing the angle in exponential by 2. And i guess it will also look much nicer.
$endgroup$
– maik
Mar 31 at 11:51
add a comment |
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$begingroup$
my main issue is applying the cosine rule to get the third side. Whats its complex equivalent?
$endgroup$
– maik
Mar 31 at 5:40
$begingroup$
What about $z_B-z_C$?
$endgroup$
– GReyes
Mar 31 at 5:45
$begingroup$
not given, but two sides and the angle in between is known, so the remaining angles and sides can be uniquely determined, at least using trigonometry i know how to find this.
$endgroup$
– maik
Mar 31 at 5:48
$begingroup$
You need $|z_B-z_C|^2=(z_B-z_C)overline(z_B-z_C)$ and $z_B-z_C=(z_B-z_A)+(z_A-z_C)$. You know that $|z_B-z_A|^2=c^2$ and $|z_C-z_A|^2=b^2$...
$endgroup$
– GReyes
Mar 31 at 5:52
$begingroup$
we also know $angle BAC = alpha$
$endgroup$
– maik
Mar 31 at 5:53