Calculate the coordinates of the third vertex of triangle given the other two and the length of edges in the...
I simply have a triangle.. say abc . Given coordinates of a & b. and the length of
ac and bc..
I can calculate the length between ab via the distance square rule.
$D = sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2} $
I have tried to use the distance rule to compute the third vertex, but it has a lot of steps..
I wonder if there is another method.. that yields the two possible solutions for the vertex
trigonometry triangle
asked Feb 22 '17 at 21:01
Shady Atef
1135
add a comment |
I simply have a triangle.. say abc . Given coordinates of a & b. and the length of
ac and bc..
I can calculate the length between ab via the distance square rule.
$D = sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2} $
I have tried to use the distance rule to compute the third vertex, but it has a lot of steps..
I wonder if there is another method.. that yields the two possible solutions for the vertex
trigonometry triangle
asked Feb 22 '17 at 21:01
Shady Atef
1135
What are the X and Y? I don't quite understand what you're asking here...
– The Count
Feb 22 '17 at 21:02
X and Y are just place holders for the coordinates of any point in the Cartesian plane.. and D is the distance between them
– Shady Atef
Feb 22 '17 at 21:03
add a comment |
I simply have a triangle.. say abc . Given coordinates of a & b. and the length of
ac and bc..
I can calculate the length between ab via the distance square rule.
$D = sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2} $
I have tried to use the distance rule to compute the third vertex, but it has a lot of steps..
I wonder if there is another method.. that yields the two possible solutions for the vertex
trigonometry triangle
asked Feb 22 '17 at 21:01
Shady Atef
1135
I simply have a triangle.. say abc . Given coordinates of a & b. and the length of
ac and bc..
I can calculate the length between ab via the distance square rule.
$D = sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2} $
I have tried to use the distance rule to compute the third vertex, but it has a lot of steps..
I wonder if there is another method.. that yields the two possible solutions for the vertex
trigonometry triangle
trigonometry triangle
asked Feb 22 '17 at 21:01
Shady Atef
1135
asked Feb 22 '17 at 21:01
Shady Atef
1135
asked Feb 22 '17 at 21:01
Shady Atef
1135
asked Feb 22 '17 at 21:01
Shady Atef
1135
asked Feb 22 '17 at 21:01
Shady Atef
1135
1135
What are the X and Y? I don't quite understand what you're asking here...
– The Count
Feb 22 '17 at 21:02
X and Y are just place holders for the coordinates of any point in the Cartesian plane.. and D is the distance between them
– Shady Atef
Feb 22 '17 at 21:03
add a comment |
What are the X and Y? I don't quite understand what you're asking here...
– The Count
Feb 22 '17 at 21:02
X and Y are just place holders for the coordinates of any point in the Cartesian plane.. and D is the distance between them
– Shady Atef
Feb 22 '17 at 21:03
What are the X and Y? I don't quite understand what you're asking here...
– The Count
Feb 22 '17 at 21:02
What are the X and Y? I don't quite understand what you're asking here...
– The Count
Feb 22 '17 at 21:02
X and Y are just place holders for the coordinates of any point in the Cartesian plane.. and D is the distance between them
– Shady Atef
Feb 22 '17 at 21:03
X and Y are just place holders for the coordinates of any point in the Cartesian plane.. and D is the distance between them
– Shady Atef
Feb 22 '17 at 21:03
add a comment |
2 Answers
2
active
oldest
votes
Let $A = (x_A, y_A)$ and $B = (x_B,y_B)$ the known vertices of your triangle. Let's call $d_{AB}$, $d_{BC}$ and $d_{CA}$ the lengths of each side.
- Translate your points subtracting $x_A$ and $y_A$ so that $A$ corresponds with the origin. That is:
$$A' = (0, 0), B' = (x_B-x_A, y_B-y_A ) = (x_B', y_B').$$
- Rotate $B'$ so that it lies on the $x$-axis. This can be done without knowing the angle, indeed:
$$A'' = (0,0), B'' = (d_{AB}, 0).$$
Anyway, the value of the rotation angle is important for the next steps. In particular it is $$theta = arctan2left(y_B-y_A,x_B-x_Aright),$$
where $arctan2(cdot, cdot)$ is defined in details here.
- At this point, it is easy to find $C''$. Notice that there are two solutions, since the point $C''$ can be placed above or below the side $AB$.
$$x_C'' = frac{d_{AB}^2+d_{AC}^2-d_{BC}^2}{2d_{AB}},$$
and
$$y_C'' = pmfrac{sqrt{(d_{AB}+d_{AC}+d_{BC})(d_{AB}+d_{AC}-d_{BC})(d_{AB}-d_{AC}+d_{BC})(-d_{AB}+d_{AC}+d_{BC})}}{2d_{AB}}.$$
- Now, rotate back your point $C''$ using $-theta$ (see step 2), thus obtaining $C'$.
- Finally, translate $C'$ by adding $x_A$ and $y_A$ to the components in order to obtain $C$.
answered Feb 22 '17 at 21:40
the_candyman
8,73122044
2
I was astonished .. Thanks
– Shady Atef
Feb 22 '17 at 22:02
@ShadyAtef you are welcome
– the_candyman
Feb 22 '17 at 22:03
It would be nice if you also provide the way to rotate C'' in step 4.
– DDRamone
Apr 24 '18 at 13:29
This was a massive help, thank you. A few post-implementation notes: In step 2, calculateθ = arctan2(B'y, B'x)as it's simpler. Thus calculated, θ is the angle from the x-axis to B', so it is actually the opposite of the triangle's rotation angle as stated. So don't negate θ for step 4. Step 4, to counter-rotate your two C'' points [call them C1'' & C2''] back to C1' and C2' do:C1'x = C''x * cos(θ) - C1''y * sin(θ),C1'y = C''x * sin(θ) + C1''y * cos(0)andC2'x = C''x * cos(θ) - C2''y * sin(θ),C2'y = C''x * sin(θ) + C2''y * cos(0).
– par
Jul 19 '18 at 5:02
add a comment |
There is much easier solution, which has 3 steps instead of 5 and doesn't require translating or rotating:
$φ_1 = arctan2(B_y - A_y, B_x - A_x)$
$φ_2 = arccosleft(dfrac{l_1^2 + l_3^2 - l_2^2}{2cdot l_1cdot l_3}right) $
$C = A + l_1cdot[cos(φ_1±φ_2)$; $sin(φ_1±φ_2)]$
Where $A_x$, $A_y$, $B_x$, $B_y$ are your given coordinates and $l_1$, $l_2$, $l_3$ are lengths of $AC$, $BC$ and $AB$ respectively (see the image). Note that there is $±$ sign, because you can build 2 triangles which will satisfy your problem.
Problem legend
This answer may be a bit late, but I hope other people who face this problem will find my solution useful.
edited Dec 1 '18 at 13:40
Tianlalu
3,09621038
answered Dec 1 '18 at 13:30
Naiten
134
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $A = (x_A, y_A)$ and $B = (x_B,y_B)$ the known vertices of your triangle. Let's call $d_{AB}$, $d_{BC}$ and $d_{CA}$ the lengths of each side.
- Translate your points subtracting $x_A$ and $y_A$ so that $A$ corresponds with the origin. That is:
$$A' = (0, 0), B' = (x_B-x_A, y_B-y_A ) = (x_B', y_B').$$
- Rotate $B'$ so that it lies on the $x$-axis. This can be done without knowing the angle, indeed:
$$A'' = (0,0), B'' = (d_{AB}, 0).$$
Anyway, the value of the rotation angle is important for the next steps. In particular it is $$theta = arctan2left(y_B-y_A,x_B-x_Aright),$$
where $arctan2(cdot, cdot)$ is defined in details here.
- At this point, it is easy to find $C''$. Notice that there are two solutions, since the point $C''$ can be placed above or below the side $AB$.
$$x_C'' = frac{d_{AB}^2+d_{AC}^2-d_{BC}^2}{2d_{AB}},$$
and
$$y_C'' = pmfrac{sqrt{(d_{AB}+d_{AC}+d_{BC})(d_{AB}+d_{AC}-d_{BC})(d_{AB}-d_{AC}+d_{BC})(-d_{AB}+d_{AC}+d_{BC})}}{2d_{AB}}.$$
- Now, rotate back your point $C''$ using $-theta$ (see step 2), thus obtaining $C'$.
- Finally, translate $C'$ by adding $x_A$ and $y_A$ to the components in order to obtain $C$.
answered Feb 22 '17 at 21:40
the_candyman
8,73122044
2
I was astonished .. Thanks
– Shady Atef
Feb 22 '17 at 22:02
@ShadyAtef you are welcome
– the_candyman
Feb 22 '17 at 22:03
It would be nice if you also provide the way to rotate C'' in step 4.
– DDRamone
Apr 24 '18 at 13:29
This was a massive help, thank you. A few post-implementation notes: In step 2, calculateθ = arctan2(B'y, B'x)as it's simpler. Thus calculated, θ is the angle from the x-axis to B', so it is actually the opposite of the triangle's rotation angle as stated. So don't negate θ for step 4. Step 4, to counter-rotate your two C'' points [call them C1'' & C2''] back to C1' and C2' do:C1'x = C''x * cos(θ) - C1''y * sin(θ),C1'y = C''x * sin(θ) + C1''y * cos(0)andC2'x = C''x * cos(θ) - C2''y * sin(θ),C2'y = C''x * sin(θ) + C2''y * cos(0).
– par
Jul 19 '18 at 5:02
add a comment |
Let $A = (x_A, y_A)$ and $B = (x_B,y_B)$ the known vertices of your triangle. Let's call $d_{AB}$, $d_{BC}$ and $d_{CA}$ the lengths of each side.
- Translate your points subtracting $x_A$ and $y_A$ so that $A$ corresponds with the origin. That is:
$$A' = (0, 0), B' = (x_B-x_A, y_B-y_A ) = (x_B', y_B').$$
- Rotate $B'$ so that it lies on the $x$-axis. This can be done without knowing the angle, indeed:
$$A'' = (0,0), B'' = (d_{AB}, 0).$$
Anyway, the value of the rotation angle is important for the next steps. In particular it is $$theta = arctan2left(y_B-y_A,x_B-x_Aright),$$
where $arctan2(cdot, cdot)$ is defined in details here.
- At this point, it is easy to find $C''$. Notice that there are two solutions, since the point $C''$ can be placed above or below the side $AB$.
$$x_C'' = frac{d_{AB}^2+d_{AC}^2-d_{BC}^2}{2d_{AB}},$$
and
$$y_C'' = pmfrac{sqrt{(d_{AB}+d_{AC}+d_{BC})(d_{AB}+d_{AC}-d_{BC})(d_{AB}-d_{AC}+d_{BC})(-d_{AB}+d_{AC}+d_{BC})}}{2d_{AB}}.$$
- Now, rotate back your point $C''$ using $-theta$ (see step 2), thus obtaining $C'$.
- Finally, translate $C'$ by adding $x_A$ and $y_A$ to the components in order to obtain $C$.
answered Feb 22 '17 at 21:40
the_candyman
8,73122044
2
I was astonished .. Thanks
– Shady Atef
Feb 22 '17 at 22:02
@ShadyAtef you are welcome
– the_candyman
Feb 22 '17 at 22:03
It would be nice if you also provide the way to rotate C'' in step 4.
– DDRamone
Apr 24 '18 at 13:29
This was a massive help, thank you. A few post-implementation notes: In step 2, calculateθ = arctan2(B'y, B'x)as it's simpler. Thus calculated, θ is the angle from the x-axis to B', so it is actually the opposite of the triangle's rotation angle as stated. So don't negate θ for step 4. Step 4, to counter-rotate your two C'' points [call them C1'' & C2''] back to C1' and C2' do:C1'x = C''x * cos(θ) - C1''y * sin(θ),C1'y = C''x * sin(θ) + C1''y * cos(0)andC2'x = C''x * cos(θ) - C2''y * sin(θ),C2'y = C''x * sin(θ) + C2''y * cos(0).
– par
Jul 19 '18 at 5:02
add a comment |
Let $A = (x_A, y_A)$ and $B = (x_B,y_B)$ the known vertices of your triangle. Let's call $d_{AB}$, $d_{BC}$ and $d_{CA}$ the lengths of each side.
- Translate your points subtracting $x_A$ and $y_A$ so that $A$ corresponds with the origin. That is:
$$A' = (0, 0), B' = (x_B-x_A, y_B-y_A ) = (x_B', y_B').$$
- Rotate $B'$ so that it lies on the $x$-axis. This can be done without knowing the angle, indeed:
$$A'' = (0,0), B'' = (d_{AB}, 0).$$
Anyway, the value of the rotation angle is important for the next steps. In particular it is $$theta = arctan2left(y_B-y_A,x_B-x_Aright),$$
where $arctan2(cdot, cdot)$ is defined in details here.
- At this point, it is easy to find $C''$. Notice that there are two solutions, since the point $C''$ can be placed above or below the side $AB$.
$$x_C'' = frac{d_{AB}^2+d_{AC}^2-d_{BC}^2}{2d_{AB}},$$
and
$$y_C'' = pmfrac{sqrt{(d_{AB}+d_{AC}+d_{BC})(d_{AB}+d_{AC}-d_{BC})(d_{AB}-d_{AC}+d_{BC})(-d_{AB}+d_{AC}+d_{BC})}}{2d_{AB}}.$$
- Now, rotate back your point $C''$ using $-theta$ (see step 2), thus obtaining $C'$.
- Finally, translate $C'$ by adding $x_A$ and $y_A$ to the components in order to obtain $C$.
answered Feb 22 '17 at 21:40
the_candyman
8,73122044
Let $A = (x_A, y_A)$ and $B = (x_B,y_B)$ the known vertices of your triangle. Let's call $d_{AB}$, $d_{BC}$ and $d_{CA}$ the lengths of each side.
- Translate your points subtracting $x_A$ and $y_A$ so that $A$ corresponds with the origin. That is:
$$A' = (0, 0), B' = (x_B-x_A, y_B-y_A ) = (x_B', y_B').$$
- Rotate $B'$ so that it lies on the $x$-axis. This can be done without knowing the angle, indeed:
$$A'' = (0,0), B'' = (d_{AB}, 0).$$
Anyway, the value of the rotation angle is important for the next steps. In particular it is $$theta = arctan2left(y_B-y_A,x_B-x_Aright),$$
where $arctan2(cdot, cdot)$ is defined in details here.
- At this point, it is easy to find $C''$. Notice that there are two solutions, since the point $C''$ can be placed above or below the side $AB$.
$$x_C'' = frac{d_{AB}^2+d_{AC}^2-d_{BC}^2}{2d_{AB}},$$
and
$$y_C'' = pmfrac{sqrt{(d_{AB}+d_{AC}+d_{BC})(d_{AB}+d_{AC}-d_{BC})(d_{AB}-d_{AC}+d_{BC})(-d_{AB}+d_{AC}+d_{BC})}}{2d_{AB}}.$$
- Now, rotate back your point $C''$ using $-theta$ (see step 2), thus obtaining $C'$.
- Finally, translate $C'$ by adding $x_A$ and $y_A$ to the components in order to obtain $C$.
answered Feb 22 '17 at 21:40
the_candyman
8,73122044
answered Feb 22 '17 at 21:40
the_candyman
8,73122044
answered Feb 22 '17 at 21:40
the_candyman
8,73122044
answered Feb 22 '17 at 21:40
the_candyman
8,73122044
8,73122044
2
I was astonished .. Thanks
– Shady Atef
Feb 22 '17 at 22:02
@ShadyAtef you are welcome
– the_candyman
Feb 22 '17 at 22:03
It would be nice if you also provide the way to rotate C'' in step 4.
– DDRamone
Apr 24 '18 at 13:29
This was a massive help, thank you. A few post-implementation notes: In step 2, calculateθ = arctan2(B'y, B'x)as it's simpler. Thus calculated, θ is the angle from the x-axis to B', so it is actually the opposite of the triangle's rotation angle as stated. So don't negate θ for step 4. Step 4, to counter-rotate your two C'' points [call them C1'' & C2''] back to C1' and C2' do:C1'x = C''x * cos(θ) - C1''y * sin(θ),C1'y = C''x * sin(θ) + C1''y * cos(0)andC2'x = C''x * cos(θ) - C2''y * sin(θ),C2'y = C''x * sin(θ) + C2''y * cos(0).
– par
Jul 19 '18 at 5:02
add a comment |
2
I was astonished .. Thanks
– Shady Atef
Feb 22 '17 at 22:02
@ShadyAtef you are welcome
– the_candyman
Feb 22 '17 at 22:03
It would be nice if you also provide the way to rotate C'' in step 4.
– DDRamone
Apr 24 '18 at 13:29
This was a massive help, thank you. A few post-implementation notes: In step 2, calculateθ = arctan2(B'y, B'x)as it's simpler. Thus calculated, θ is the angle from the x-axis to B', so it is actually the opposite of the triangle's rotation angle as stated. So don't negate θ for step 4. Step 4, to counter-rotate your two C'' points [call them C1'' & C2''] back to C1' and C2' do:C1'x = C''x * cos(θ) - C1''y * sin(θ),C1'y = C''x * sin(θ) + C1''y * cos(0)andC2'x = C''x * cos(θ) - C2''y * sin(θ),C2'y = C''x * sin(θ) + C2''y * cos(0).
– par
Jul 19 '18 at 5:02
2
2
I was astonished .. Thanks
– Shady Atef
Feb 22 '17 at 22:02
I was astonished .. Thanks
– Shady Atef
Feb 22 '17 at 22:02
@ShadyAtef you are welcome
– the_candyman
Feb 22 '17 at 22:03
@ShadyAtef you are welcome
– the_candyman
Feb 22 '17 at 22:03
It would be nice if you also provide the way to rotate C'' in step 4.
– DDRamone
Apr 24 '18 at 13:29
It would be nice if you also provide the way to rotate C'' in step 4.
– DDRamone
Apr 24 '18 at 13:29
This was a massive help, thank you. A few post-implementation notes: In step 2, calculate
θ = arctan2(B'y, B'x) as it's simpler. Thus calculated, θ is the angle from the x-axis to B', so it is actually the opposite of the triangle's rotation angle as stated. So don't negate θ for step 4. Step 4, to counter-rotate your two C'' points [call them C1'' & C2''] back to C1' and C2' do: C1'x = C''x * cos(θ) - C1''y * sin(θ), C1'y = C''x * sin(θ) + C1''y * cos(0) and C2'x = C''x * cos(θ) - C2''y * sin(θ), C2'y = C''x * sin(θ) + C2''y * cos(0).– par
Jul 19 '18 at 5:02
This was a massive help, thank you. A few post-implementation notes: In step 2, calculate
θ = arctan2(B'y, B'x) as it's simpler. Thus calculated, θ is the angle from the x-axis to B', so it is actually the opposite of the triangle's rotation angle as stated. So don't negate θ for step 4. Step 4, to counter-rotate your two C'' points [call them C1'' & C2''] back to C1' and C2' do: C1'x = C''x * cos(θ) - C1''y * sin(θ), C1'y = C''x * sin(θ) + C1''y * cos(0) and C2'x = C''x * cos(θ) - C2''y * sin(θ), C2'y = C''x * sin(θ) + C2''y * cos(0).– par
Jul 19 '18 at 5:02
add a comment |
There is much easier solution, which has 3 steps instead of 5 and doesn't require translating or rotating:
$φ_1 = arctan2(B_y - A_y, B_x - A_x)$
$φ_2 = arccosleft(dfrac{l_1^2 + l_3^2 - l_2^2}{2cdot l_1cdot l_3}right) $
$C = A + l_1cdot[cos(φ_1±φ_2)$; $sin(φ_1±φ_2)]$
Where $A_x$, $A_y$, $B_x$, $B_y$ are your given coordinates and $l_1$, $l_2$, $l_3$ are lengths of $AC$, $BC$ and $AB$ respectively (see the image). Note that there is $±$ sign, because you can build 2 triangles which will satisfy your problem.
Problem legend
This answer may be a bit late, but I hope other people who face this problem will find my solution useful.
edited Dec 1 '18 at 13:40
Tianlalu
3,09621038
answered Dec 1 '18 at 13:30
Naiten
134
add a comment |
There is much easier solution, which has 3 steps instead of 5 and doesn't require translating or rotating:
$φ_1 = arctan2(B_y - A_y, B_x - A_x)$
$φ_2 = arccosleft(dfrac{l_1^2 + l_3^2 - l_2^2}{2cdot l_1cdot l_3}right) $
$C = A + l_1cdot[cos(φ_1±φ_2)$; $sin(φ_1±φ_2)]$
Where $A_x$, $A_y$, $B_x$, $B_y$ are your given coordinates and $l_1$, $l_2$, $l_3$ are lengths of $AC$, $BC$ and $AB$ respectively (see the image). Note that there is $±$ sign, because you can build 2 triangles which will satisfy your problem.
Problem legend
This answer may be a bit late, but I hope other people who face this problem will find my solution useful.
edited Dec 1 '18 at 13:40
Tianlalu
3,09621038
answered Dec 1 '18 at 13:30
Naiten
134
add a comment |
There is much easier solution, which has 3 steps instead of 5 and doesn't require translating or rotating:
$φ_1 = arctan2(B_y - A_y, B_x - A_x)$
$φ_2 = arccosleft(dfrac{l_1^2 + l_3^2 - l_2^2}{2cdot l_1cdot l_3}right) $
$C = A + l_1cdot[cos(φ_1±φ_2)$; $sin(φ_1±φ_2)]$
Where $A_x$, $A_y$, $B_x$, $B_y$ are your given coordinates and $l_1$, $l_2$, $l_3$ are lengths of $AC$, $BC$ and $AB$ respectively (see the image). Note that there is $±$ sign, because you can build 2 triangles which will satisfy your problem.
Problem legend
This answer may be a bit late, but I hope other people who face this problem will find my solution useful.
edited Dec 1 '18 at 13:40
Tianlalu
3,09621038
answered Dec 1 '18 at 13:30
Naiten
134
There is much easier solution, which has 3 steps instead of 5 and doesn't require translating or rotating:
$φ_1 = arctan2(B_y - A_y, B_x - A_x)$
$φ_2 = arccosleft(dfrac{l_1^2 + l_3^2 - l_2^2}{2cdot l_1cdot l_3}right) $
$C = A + l_1cdot[cos(φ_1±φ_2)$; $sin(φ_1±φ_2)]$
Where $A_x$, $A_y$, $B_x$, $B_y$ are your given coordinates and $l_1$, $l_2$, $l_3$ are lengths of $AC$, $BC$ and $AB$ respectively (see the image). Note that there is $±$ sign, because you can build 2 triangles which will satisfy your problem.
Problem legend
This answer may be a bit late, but I hope other people who face this problem will find my solution useful.
edited Dec 1 '18 at 13:40
Tianlalu
3,09621038
answered Dec 1 '18 at 13:30
Naiten
134
edited Dec 1 '18 at 13:40
Tianlalu
3,09621038
edited Dec 1 '18 at 13:40
Tianlalu
3,09621038
edited Dec 1 '18 at 13:40
Tianlalu
3,09621038
3,09621038
answered Dec 1 '18 at 13:30
Naiten
134
answered Dec 1 '18 at 13:30
Naiten
134
answered Dec 1 '18 at 13:30
Naiten
134
134
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What are the X and Y? I don't quite understand what you're asking here...
– The Count
Feb 22 '17 at 21:02
X and Y are just place holders for the coordinates of any point in the Cartesian plane.. and D is the distance between them
– Shady Atef
Feb 22 '17 at 21:03