How to plot a triangle given three side lengths?
$begingroup$
I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
I tried the solution from the other topic and it didn't work since the result produced was C(20,0) and that cant be right since one of the sides is already placed on the axis and its not a right triangle.
Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?
Thank you.
geometry trigonometry triangle
asked Dec 18 '18 at 20:42
Stand StudiosStand Studios
1
$endgroup$
add a comment |
$begingroup$
I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
I tried the solution from the other topic and it didn't work since the result produced was C(20,0) and that cant be right since one of the sides is already placed on the axis and its not a right triangle.
Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?
Thank you.
geometry trigonometry triangle
asked Dec 18 '18 at 20:42
Stand StudiosStand Studios
1
$endgroup$
6
$begingroup$
There is no triangle with sides such that $a+b=c$.
$endgroup$
– AugSB
Dec 18 '18 at 20:48
1
$begingroup$
It gave you that answer because that was the correct answer. A triangle with $a= 10, b = 20, c = 30$ is not a right triangle because $10^2 + 20^2 ne 30^2$. A triangle with side $a=10, b=20, c=30$ is not a triangle at all but an straight line segment.
$endgroup$
– fleablood
Dec 18 '18 at 20:54
$begingroup$
You can read up on Triangle Inequality if you want... It seems that sometimes we refer to this case as a degenerate triangle which I think of (and seems to be the common parlance) as more of a line segment than a triangle.
$endgroup$
– Mason
Dec 18 '18 at 21:00
add a comment |
$begingroup$
I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
I tried the solution from the other topic and it didn't work since the result produced was C(20,0) and that cant be right since one of the sides is already placed on the axis and its not a right triangle.
Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?
Thank you.
geometry trigonometry triangle
asked Dec 18 '18 at 20:42
Stand StudiosStand Studios
1
$endgroup$
I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
I tried the solution from the other topic and it didn't work since the result produced was C(20,0) and that cant be right since one of the sides is already placed on the axis and its not a right triangle.
Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?
Thank you.
geometry trigonometry triangle
geometry trigonometry triangle
asked Dec 18 '18 at 20:42
Stand StudiosStand Studios
1
asked Dec 18 '18 at 20:42
Stand StudiosStand Studios
1
asked Dec 18 '18 at 20:42
Stand StudiosStand Studios
1
asked Dec 18 '18 at 20:42
Stand StudiosStand Studios
1
asked Dec 18 '18 at 20:42
Stand StudiosStand Studios
1
1
6
$begingroup$
There is no triangle with sides such that $a+b=c$.
$endgroup$
– AugSB
Dec 18 '18 at 20:48
1
$begingroup$
It gave you that answer because that was the correct answer. A triangle with $a= 10, b = 20, c = 30$ is not a right triangle because $10^2 + 20^2 ne 30^2$. A triangle with side $a=10, b=20, c=30$ is not a triangle at all but an straight line segment.
$endgroup$
– fleablood
Dec 18 '18 at 20:54
$begingroup$
You can read up on Triangle Inequality if you want... It seems that sometimes we refer to this case as a degenerate triangle which I think of (and seems to be the common parlance) as more of a line segment than a triangle.
$endgroup$
– Mason
Dec 18 '18 at 21:00
add a comment |
6
$begingroup$
There is no triangle with sides such that $a+b=c$.
$endgroup$
– AugSB
Dec 18 '18 at 20:48
1
$begingroup$
It gave you that answer because that was the correct answer. A triangle with $a= 10, b = 20, c = 30$ is not a right triangle because $10^2 + 20^2 ne 30^2$. A triangle with side $a=10, b=20, c=30$ is not a triangle at all but an straight line segment.
$endgroup$
– fleablood
Dec 18 '18 at 20:54
$begingroup$
You can read up on Triangle Inequality if you want... It seems that sometimes we refer to this case as a degenerate triangle which I think of (and seems to be the common parlance) as more of a line segment than a triangle.
$endgroup$
– Mason
Dec 18 '18 at 21:00
6
6
$begingroup$
There is no triangle with sides such that $a+b=c$.
$endgroup$
– AugSB
Dec 18 '18 at 20:48
$begingroup$
There is no triangle with sides such that $a+b=c$.
$endgroup$
– AugSB
Dec 18 '18 at 20:48
1
1
$begingroup$
It gave you that answer because that was the correct answer. A triangle with $a= 10, b = 20, c = 30$ is not a right triangle because $10^2 + 20^2 ne 30^2$. A triangle with side $a=10, b=20, c=30$ is not a triangle at all but an straight line segment.
$endgroup$
– fleablood
Dec 18 '18 at 20:54
$begingroup$
It gave you that answer because that was the correct answer. A triangle with $a= 10, b = 20, c = 30$ is not a right triangle because $10^2 + 20^2 ne 30^2$. A triangle with side $a=10, b=20, c=30$ is not a triangle at all but an straight line segment.
$endgroup$
– fleablood
Dec 18 '18 at 20:54
$begingroup$
You can read up on Triangle Inequality if you want... It seems that sometimes we refer to this case as a degenerate triangle which I think of (and seems to be the common parlance) as more of a line segment than a triangle.
$endgroup$
– Mason
Dec 18 '18 at 21:00
$begingroup$
You can read up on Triangle Inequality if you want... It seems that sometimes we refer to this case as a degenerate triangle which I think of (and seems to be the common parlance) as more of a line segment than a triangle.
$endgroup$
– Mason
Dec 18 '18 at 21:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
By the triangle inequality $a + b > c$. In this case you have $a + b =c$. So this will be a straight line where $C$ is in a line in between $A$ and $B$.
I tried the solution from the other topic
WHAT other topic?
and it didn't work since the result produced was C(20,0)
That is correct. $A = (0,0)$ and $B = (30, 0)$ and $C = (20,0)$ then $a = BC = sqrt{(30-20)^2 - (0-0)^2} = sqrt{(10)^2} = 10$ and $b = AC = sqrt{(0-20)^2 - (0-0)^2} = sqrt{(-20)^2} = 20$ and $c = AB = sqrt{(0-30)^2 - (0-0)^2} = sqrt{ (-30)^2} = 30$.
and that cant be right since one of the sides is already placed on the axis and its not a right triangle.
Who said it was a right triangle?
$a^2 + b^2 = 10^2 + 20^2 = 100 + 400 = 500$ but $c^2 = 30^2 = 900$ and $500 ne 900$. It isn't a right triangle.
Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?
So if $A= (0,0)$ and $B= (c,0)$ then $C = (x,y)$ where
$b^2 = AC^2 = x^2 +y^2$
$a^2 = BC^2 = (x-c)^2 + y^2$
So $a^2 - b^2 = (x-c)^2 - x^2$ or
$a^2 - b^2 = x^2 - 2cx + c^2 - y^2 = -2cx + c^2$ or
$x = frac {c^2 + b^2 -a^2}{2c}$
And so $x^2 +y^2 = b^2$ so
$y = sqrt{ b^2 - x^2}$
Plugging in $a = 10; b=20; c= 30$ we get
$x =frac {c^2 + b^2 -a^2}{2c}= frac {30^2 + 20^2 -10^2}{2*30}= frac {900+ 400 - 100}{60} = frac {1200}{60} = 20$
And $x =sqrt{ b^2 - x^2}= sqrt {20^2 - 20^2} = 0$.
So $C = (20, 0)$
For a more proper triangle sa $a = 20; b= 15; c=25$ (an actual right triangle.
You get $A=(0,0)$ and $B = (25,0)$ and $C = (x,y)$ where
$x = frac {c^2 + b^2 -a^2}{2c}= frac {25^2 + 15^2-20^2}{50}= 9 $
and
$y = sqrt{ b^2 - x^2}= sqrt{15^2 - 9^2} = 12$
So $C = (9,12)$.
======
Now for an example where that fails: Say $a = 20; b=30; c = 60$. That's impossible because $a+b < c$. No such triangle exists.
If $A=(0,0)$ and $B=(60,0)$ then $C = (x,y)$ with
$x = frac {c^2 + b^2 -a^2}{2c}= frac {60^2 + 30^2-20^2}{60}= 10sqrt{41}$
and
$y = sqrt{ b^2 - x^2}= sqrt{20^2 - (10sqrt{41}^2} = sqrt{400-410} = sqrt{-10}$
So $C = (10sqrt{41},sqrt{-10})$ is not possible.
answered Dec 18 '18 at 21:18
fleabloodfleablood
69.4k22685
$endgroup$
$begingroup$
Thank you. I guess it was some kind of trick question "In main(), create one instance each of the following: a Circle with a radius of 23, a Square with sides 25, and a Triangle with sides 10, 20, 30. Define all of them so that the origin (0,0) is somewhere within each object. Display the information from each object." I know that it's not a right triangle, I said that its not.
$endgroup$
– Stand Studios
Dec 19 '18 at 14:27
add a comment |
$begingroup$
Given sides a,b,and c, use the law of cosines to calculate the angle $theta$ between a and b. then (for example), plot b as horizontal, and plot a as a directional vector, with angle $theta.$
answered Dec 18 '18 at 23:07
user2661923user2661923
510112
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
By the triangle inequality $a + b > c$. In this case you have $a + b =c$. So this will be a straight line where $C$ is in a line in between $A$ and $B$.
I tried the solution from the other topic
WHAT other topic?
and it didn't work since the result produced was C(20,0)
That is correct. $A = (0,0)$ and $B = (30, 0)$ and $C = (20,0)$ then $a = BC = sqrt{(30-20)^2 - (0-0)^2} = sqrt{(10)^2} = 10$ and $b = AC = sqrt{(0-20)^2 - (0-0)^2} = sqrt{(-20)^2} = 20$ and $c = AB = sqrt{(0-30)^2 - (0-0)^2} = sqrt{ (-30)^2} = 30$.
and that cant be right since one of the sides is already placed on the axis and its not a right triangle.
Who said it was a right triangle?
$a^2 + b^2 = 10^2 + 20^2 = 100 + 400 = 500$ but $c^2 = 30^2 = 900$ and $500 ne 900$. It isn't a right triangle.
Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?
So if $A= (0,0)$ and $B= (c,0)$ then $C = (x,y)$ where
$b^2 = AC^2 = x^2 +y^2$
$a^2 = BC^2 = (x-c)^2 + y^2$
So $a^2 - b^2 = (x-c)^2 - x^2$ or
$a^2 - b^2 = x^2 - 2cx + c^2 - y^2 = -2cx + c^2$ or
$x = frac {c^2 + b^2 -a^2}{2c}$
And so $x^2 +y^2 = b^2$ so
$y = sqrt{ b^2 - x^2}$
Plugging in $a = 10; b=20; c= 30$ we get
$x =frac {c^2 + b^2 -a^2}{2c}= frac {30^2 + 20^2 -10^2}{2*30}= frac {900+ 400 - 100}{60} = frac {1200}{60} = 20$
And $x =sqrt{ b^2 - x^2}= sqrt {20^2 - 20^2} = 0$.
So $C = (20, 0)$
For a more proper triangle sa $a = 20; b= 15; c=25$ (an actual right triangle.
You get $A=(0,0)$ and $B = (25,0)$ and $C = (x,y)$ where
$x = frac {c^2 + b^2 -a^2}{2c}= frac {25^2 + 15^2-20^2}{50}= 9 $
and
$y = sqrt{ b^2 - x^2}= sqrt{15^2 - 9^2} = 12$
So $C = (9,12)$.
======
Now for an example where that fails: Say $a = 20; b=30; c = 60$. That's impossible because $a+b < c$. No such triangle exists.
If $A=(0,0)$ and $B=(60,0)$ then $C = (x,y)$ with
$x = frac {c^2 + b^2 -a^2}{2c}= frac {60^2 + 30^2-20^2}{60}= 10sqrt{41}$
and
$y = sqrt{ b^2 - x^2}= sqrt{20^2 - (10sqrt{41}^2} = sqrt{400-410} = sqrt{-10}$
So $C = (10sqrt{41},sqrt{-10})$ is not possible.
answered Dec 18 '18 at 21:18
fleabloodfleablood
69.4k22685
$endgroup$
$begingroup$
Thank you. I guess it was some kind of trick question "In main(), create one instance each of the following: a Circle with a radius of 23, a Square with sides 25, and a Triangle with sides 10, 20, 30. Define all of them so that the origin (0,0) is somewhere within each object. Display the information from each object." I know that it's not a right triangle, I said that its not.
$endgroup$
– Stand Studios
Dec 19 '18 at 14:27
add a comment |
$begingroup$
I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
By the triangle inequality $a + b > c$. In this case you have $a + b =c$. So this will be a straight line where $C$ is in a line in between $A$ and $B$.
I tried the solution from the other topic
WHAT other topic?
and it didn't work since the result produced was C(20,0)
That is correct. $A = (0,0)$ and $B = (30, 0)$ and $C = (20,0)$ then $a = BC = sqrt{(30-20)^2 - (0-0)^2} = sqrt{(10)^2} = 10$ and $b = AC = sqrt{(0-20)^2 - (0-0)^2} = sqrt{(-20)^2} = 20$ and $c = AB = sqrt{(0-30)^2 - (0-0)^2} = sqrt{ (-30)^2} = 30$.
and that cant be right since one of the sides is already placed on the axis and its not a right triangle.
Who said it was a right triangle?
$a^2 + b^2 = 10^2 + 20^2 = 100 + 400 = 500$ but $c^2 = 30^2 = 900$ and $500 ne 900$. It isn't a right triangle.
Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?
So if $A= (0,0)$ and $B= (c,0)$ then $C = (x,y)$ where
$b^2 = AC^2 = x^2 +y^2$
$a^2 = BC^2 = (x-c)^2 + y^2$
So $a^2 - b^2 = (x-c)^2 - x^2$ or
$a^2 - b^2 = x^2 - 2cx + c^2 - y^2 = -2cx + c^2$ or
$x = frac {c^2 + b^2 -a^2}{2c}$
And so $x^2 +y^2 = b^2$ so
$y = sqrt{ b^2 - x^2}$
Plugging in $a = 10; b=20; c= 30$ we get
$x =frac {c^2 + b^2 -a^2}{2c}= frac {30^2 + 20^2 -10^2}{2*30}= frac {900+ 400 - 100}{60} = frac {1200}{60} = 20$
And $x =sqrt{ b^2 - x^2}= sqrt {20^2 - 20^2} = 0$.
So $C = (20, 0)$
For a more proper triangle sa $a = 20; b= 15; c=25$ (an actual right triangle.
You get $A=(0,0)$ and $B = (25,0)$ and $C = (x,y)$ where
$x = frac {c^2 + b^2 -a^2}{2c}= frac {25^2 + 15^2-20^2}{50}= 9 $
and
$y = sqrt{ b^2 - x^2}= sqrt{15^2 - 9^2} = 12$
So $C = (9,12)$.
======
Now for an example where that fails: Say $a = 20; b=30; c = 60$. That's impossible because $a+b < c$. No such triangle exists.
If $A=(0,0)$ and $B=(60,0)$ then $C = (x,y)$ with
$x = frac {c^2 + b^2 -a^2}{2c}= frac {60^2 + 30^2-20^2}{60}= 10sqrt{41}$
and
$y = sqrt{ b^2 - x^2}= sqrt{20^2 - (10sqrt{41}^2} = sqrt{400-410} = sqrt{-10}$
So $C = (10sqrt{41},sqrt{-10})$ is not possible.
answered Dec 18 '18 at 21:18
fleabloodfleablood
69.4k22685
$endgroup$
$begingroup$
Thank you. I guess it was some kind of trick question "In main(), create one instance each of the following: a Circle with a radius of 23, a Square with sides 25, and a Triangle with sides 10, 20, 30. Define all of them so that the origin (0,0) is somewhere within each object. Display the information from each object." I know that it's not a right triangle, I said that its not.
$endgroup$
– Stand Studios
Dec 19 '18 at 14:27
add a comment |
$begingroup$
I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
By the triangle inequality $a + b > c$. In this case you have $a + b =c$. So this will be a straight line where $C$ is in a line in between $A$ and $B$.
I tried the solution from the other topic
WHAT other topic?
and it didn't work since the result produced was C(20,0)
That is correct. $A = (0,0)$ and $B = (30, 0)$ and $C = (20,0)$ then $a = BC = sqrt{(30-20)^2 - (0-0)^2} = sqrt{(10)^2} = 10$ and $b = AC = sqrt{(0-20)^2 - (0-0)^2} = sqrt{(-20)^2} = 20$ and $c = AB = sqrt{(0-30)^2 - (0-0)^2} = sqrt{ (-30)^2} = 30$.
and that cant be right since one of the sides is already placed on the axis and its not a right triangle.
Who said it was a right triangle?
$a^2 + b^2 = 10^2 + 20^2 = 100 + 400 = 500$ but $c^2 = 30^2 = 900$ and $500 ne 900$. It isn't a right triangle.
Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?
So if $A= (0,0)$ and $B= (c,0)$ then $C = (x,y)$ where
$b^2 = AC^2 = x^2 +y^2$
$a^2 = BC^2 = (x-c)^2 + y^2$
So $a^2 - b^2 = (x-c)^2 - x^2$ or
$a^2 - b^2 = x^2 - 2cx + c^2 - y^2 = -2cx + c^2$ or
$x = frac {c^2 + b^2 -a^2}{2c}$
And so $x^2 +y^2 = b^2$ so
$y = sqrt{ b^2 - x^2}$
Plugging in $a = 10; b=20; c= 30$ we get
$x =frac {c^2 + b^2 -a^2}{2c}= frac {30^2 + 20^2 -10^2}{2*30}= frac {900+ 400 - 100}{60} = frac {1200}{60} = 20$
And $x =sqrt{ b^2 - x^2}= sqrt {20^2 - 20^2} = 0$.
So $C = (20, 0)$
For a more proper triangle sa $a = 20; b= 15; c=25$ (an actual right triangle.
You get $A=(0,0)$ and $B = (25,0)$ and $C = (x,y)$ where
$x = frac {c^2 + b^2 -a^2}{2c}= frac {25^2 + 15^2-20^2}{50}= 9 $
and
$y = sqrt{ b^2 - x^2}= sqrt{15^2 - 9^2} = 12$
So $C = (9,12)$.
======
Now for an example where that fails: Say $a = 20; b=30; c = 60$. That's impossible because $a+b < c$. No such triangle exists.
If $A=(0,0)$ and $B=(60,0)$ then $C = (x,y)$ with
$x = frac {c^2 + b^2 -a^2}{2c}= frac {60^2 + 30^2-20^2}{60}= 10sqrt{41}$
and
$y = sqrt{ b^2 - x^2}= sqrt{20^2 - (10sqrt{41}^2} = sqrt{400-410} = sqrt{-10}$
So $C = (10sqrt{41},sqrt{-10})$ is not possible.
answered Dec 18 '18 at 21:18
fleabloodfleablood
69.4k22685
$endgroup$
I am trying to create a program for a school project where I need to plot points of a triangle given all 3 side lengths (a=10, b=20, c=30).
By the triangle inequality $a + b > c$. In this case you have $a + b =c$. So this will be a straight line where $C$ is in a line in between $A$ and $B$.
I tried the solution from the other topic
WHAT other topic?
and it didn't work since the result produced was C(20,0)
That is correct. $A = (0,0)$ and $B = (30, 0)$ and $C = (20,0)$ then $a = BC = sqrt{(30-20)^2 - (0-0)^2} = sqrt{(10)^2} = 10$ and $b = AC = sqrt{(0-20)^2 - (0-0)^2} = sqrt{(-20)^2} = 20$ and $c = AB = sqrt{(0-30)^2 - (0-0)^2} = sqrt{ (-30)^2} = 30$.
and that cant be right since one of the sides is already placed on the axis and its not a right triangle.
Who said it was a right triangle?
$a^2 + b^2 = 10^2 + 20^2 = 100 + 400 = 500$ but $c^2 = 30^2 = 900$ and $500 ne 900$. It isn't a right triangle.
Is there a formula I can plug into my program that will produce point C, given that I place A and B on x or y axis and will work with any triangle?
So if $A= (0,0)$ and $B= (c,0)$ then $C = (x,y)$ where
$b^2 = AC^2 = x^2 +y^2$
$a^2 = BC^2 = (x-c)^2 + y^2$
So $a^2 - b^2 = (x-c)^2 - x^2$ or
$a^2 - b^2 = x^2 - 2cx + c^2 - y^2 = -2cx + c^2$ or
$x = frac {c^2 + b^2 -a^2}{2c}$
And so $x^2 +y^2 = b^2$ so
$y = sqrt{ b^2 - x^2}$
Plugging in $a = 10; b=20; c= 30$ we get
$x =frac {c^2 + b^2 -a^2}{2c}= frac {30^2 + 20^2 -10^2}{2*30}= frac {900+ 400 - 100}{60} = frac {1200}{60} = 20$
And $x =sqrt{ b^2 - x^2}= sqrt {20^2 - 20^2} = 0$.
So $C = (20, 0)$
For a more proper triangle sa $a = 20; b= 15; c=25$ (an actual right triangle.
You get $A=(0,0)$ and $B = (25,0)$ and $C = (x,y)$ where
$x = frac {c^2 + b^2 -a^2}{2c}= frac {25^2 + 15^2-20^2}{50}= 9 $
and
$y = sqrt{ b^2 - x^2}= sqrt{15^2 - 9^2} = 12$
So $C = (9,12)$.
======
Now for an example where that fails: Say $a = 20; b=30; c = 60$. That's impossible because $a+b < c$. No such triangle exists.
If $A=(0,0)$ and $B=(60,0)$ then $C = (x,y)$ with
$x = frac {c^2 + b^2 -a^2}{2c}= frac {60^2 + 30^2-20^2}{60}= 10sqrt{41}$
and
$y = sqrt{ b^2 - x^2}= sqrt{20^2 - (10sqrt{41}^2} = sqrt{400-410} = sqrt{-10}$
So $C = (10sqrt{41},sqrt{-10})$ is not possible.
answered Dec 18 '18 at 21:18
fleabloodfleablood
69.4k22685
edited Dec 18 '18 at 23:27
answered Dec 18 '18 at 21:18
fleabloodfleablood
69.4k22685
answered Dec 18 '18 at 21:18
fleabloodfleablood
69.4k22685
answered Dec 18 '18 at 21:18
fleabloodfleablood
69.4k22685
69.4k22685
$begingroup$
Thank you. I guess it was some kind of trick question "In main(), create one instance each of the following: a Circle with a radius of 23, a Square with sides 25, and a Triangle with sides 10, 20, 30. Define all of them so that the origin (0,0) is somewhere within each object. Display the information from each object." I know that it's not a right triangle, I said that its not.
$endgroup$
– Stand Studios
Dec 19 '18 at 14:27
add a comment |
$begingroup$
Thank you. I guess it was some kind of trick question "In main(), create one instance each of the following: a Circle with a radius of 23, a Square with sides 25, and a Triangle with sides 10, 20, 30. Define all of them so that the origin (0,0) is somewhere within each object. Display the information from each object." I know that it's not a right triangle, I said that its not.
$endgroup$
– Stand Studios
Dec 19 '18 at 14:27
$begingroup$
Thank you. I guess it was some kind of trick question "In main(), create one instance each of the following: a Circle with a radius of 23, a Square with sides 25, and a Triangle with sides 10, 20, 30. Define all of them so that the origin (0,0) is somewhere within each object. Display the information from each object." I know that it's not a right triangle, I said that its not.
$endgroup$
– Stand Studios
Dec 19 '18 at 14:27
$begingroup$
Thank you. I guess it was some kind of trick question "In main(), create one instance each of the following: a Circle with a radius of 23, a Square with sides 25, and a Triangle with sides 10, 20, 30. Define all of them so that the origin (0,0) is somewhere within each object. Display the information from each object." I know that it's not a right triangle, I said that its not.
$endgroup$
– Stand Studios
Dec 19 '18 at 14:27
add a comment |
$begingroup$
Given sides a,b,and c, use the law of cosines to calculate the angle $theta$ between a and b. then (for example), plot b as horizontal, and plot a as a directional vector, with angle $theta.$
answered Dec 18 '18 at 23:07
user2661923user2661923
510112
$endgroup$
add a comment |
$begingroup$
Given sides a,b,and c, use the law of cosines to calculate the angle $theta$ between a and b. then (for example), plot b as horizontal, and plot a as a directional vector, with angle $theta.$
answered Dec 18 '18 at 23:07
user2661923user2661923
510112
$endgroup$
add a comment |
$begingroup$
Given sides a,b,and c, use the law of cosines to calculate the angle $theta$ between a and b. then (for example), plot b as horizontal, and plot a as a directional vector, with angle $theta.$
answered Dec 18 '18 at 23:07
user2661923user2661923
510112
$endgroup$
Given sides a,b,and c, use the law of cosines to calculate the angle $theta$ between a and b. then (for example), plot b as horizontal, and plot a as a directional vector, with angle $theta.$
answered Dec 18 '18 at 23:07
user2661923user2661923
510112
answered Dec 18 '18 at 23:07
user2661923user2661923
510112
answered Dec 18 '18 at 23:07
user2661923user2661923
510112
answered Dec 18 '18 at 23:07
user2661923user2661923
510112
510112
add a comment |
add a comment |
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$begingroup$
There is no triangle with sides such that $a+b=c$.
$endgroup$
– AugSB
Dec 18 '18 at 20:48
1
$begingroup$
It gave you that answer because that was the correct answer. A triangle with $a= 10, b = 20, c = 30$ is not a right triangle because $10^2 + 20^2 ne 30^2$. A triangle with side $a=10, b=20, c=30$ is not a triangle at all but an straight line segment.
$endgroup$
– fleablood
Dec 18 '18 at 20:54
$begingroup$
You can read up on Triangle Inequality if you want... It seems that sometimes we refer to this case as a degenerate triangle which I think of (and seems to be the common parlance) as more of a line segment than a triangle.
$endgroup$
– Mason
Dec 18 '18 at 21:00