find the length of a triangle

[meenabala]

Hi,

please take a look at the triangle posted on the image. I need to solve x2. Is there a way to solve x2 with the variables data provided?


Thanks in advance

Meena Bala

 

[lev888]

Hi,

please take a look at the triangle posted on the image. I need to solve x2. Is there a way to solve x2 with the variables data provided?View attachment 27326


Thanks in advance

Meena Bala

I don't think we have enough information. Based on the diagram we can change x₂ while preserving the given constraints.

 

[nasi112]

I don't think we have enough information. Based on the diagram we can change x₂ while preserving the given constraints.

Dr.Peterson likes these kinds of problems which have tons of unknowns.

Where are you Dr.??

I am sure that he will figure it out

 

[Dr.Peterson]

Dr.Peterson likes these kinds of problems which have tons of unknowns.

Why would you think that?

please take a look at the triangle posted on the image. I need to solve x2. Is there a way to solve x2 with the variables data provided?

My first thought is that this is all about coordinates, so draw it out that way:



I could put A anywhere I want along that red line. This is just what lev888 said.

Do you have any additional requirements?

 

[nasi112]

Why would you think that?



My first thought is that this is all about coordinates, so draw it out that way:

View attachment 27337

I could put A anywhere I want along that red line. This is just what lev888 said.

Do you have any additional requirements?

I just feel you can solve it. Since no additional information is given, why don't you move [MATH]A[/MATH] to [MATH]40[/MATH] or [MATH]50[/MATH], and you're done.

 

[meenabala]

Why would you think that?



My first thought is that this is all about coordinates, so draw it out that way:

View attachment 27337

I could put A anywhere I want along that red line. This is just what lev888 said.

Do you have any additional requirements?

Thanks for the prompt response. I don't have any other info. But can we solve this as an equation? we have three right angled triangle if you box that scalene triangle in a rectangle and substitute those values in one of the equations for scalene triangle?

or is there any approach to calculate the angle between line (0,0) to the intersection of the vertical line(h in my diagram) from the top and then extending that triangle out to point A(as marked in your diagram)? Becasue the angle of that right angled triangle remains the same even if we extend that side out to point A ? may be it can be solved as a differential equation ?

 

[Dr.Peterson]

Thanks for the prompt response. I don't have any other info. But can we solve this as an equation? we have three right angled triangle if you box that scalene triangle in a rectangle and substitute those values in one of the equations for scalene triangle?

You seem to be totally missing the point. Without additional information, there is no one correct answer. ANY positive number could be a valid answer!

Please observe: The picture above showed one "solution"; this shows another:


I just randomly picked another location for A, which gives a different value for x_2. Both, as far as I can tell, are valid.

Now, if there were some one other fact that had to be true (maybe a specific value for the angle at A, or the length of one of the slanted lines), then there could be a unique solution you could solve an equation for. But as it is, unless you can point out something wrong with either of my pictures, there is nothing to solve.

 

[meenabala]

You seem to be totally missing the point. Without additional information, there is no one correct answer. ANY positive number could be a valid answer!

Please observe: The picture above showed one "solution"; this shows another:
View attachment 27346

I just randomly picked another location for A, which gives a different value for x_2. Both, as far as I can tell, are valid.

Now, if there were some one other fact that had to be true (maybe a specific value for the angle at A, or the length of one of the slanted lines), then there could be a unique solution you could solve an equation for. But as it is, unless you can point out something wrong with either of my pictures, there is nothing to solve.

Hello Dr. Peterson,

appreciate you looking into this. May be I did a poor job of explaining. But here's how those lines get built...let me give you an actual application; may be that will give you additional info.

Basically the line from (0,0) to (25,200) gets built first. Assume that we have number of days on the X-axis. so the line from (0,0) to (25,200) get built first in 25 days. Then the line from the top to point A (the y-coordinate of which will always be at 0.618 x 200 ) gets built few points down every subsequent days, until it gets to A. But the initial angle at which that line from (25,200) start out building, maintains that until it gets to point A. So for example, after reaching a top at 200 in 25 days, the next 10days (so 35th day from the start (0,0)) it reaches 150. So the line at 35th day will have a coordinates of (35,150). So for that line to be a side of a triangle it has to maintain the same angle right? So my question is assuming it hit 150 on 35th day, how many days would it take to reach point A( where the y-coordinate of A is always 0.618 x 200)?

 

[lev888]

Hello Dr. Peterson,

appreciate you looking into this. May be I did a poor job of explaining. But here's how those lines get built...let me give you an actual application; may be that will give you additional info.

Basically the line from (0,0) to (25,200) gets built first. Assume that we have number of days on the X-axis. so the line from (0,0) to (25,200) get built first in 25 days. Then the line from the top to point A (the y-coordinate of which will always be at 0.618 x 200 ) gets built few points down every subsequent days, until it gets to A. But the initial angle at which that line from (25,200) start out building, maintains that until it gets to point A. So for example, after reaching a top at 200 in 25 days, the next 10days (so 35th day from the start (0,0)) it reaches 150. So the line at 35th day will have a coordinates of (35,150). So for that line to be a side of a triangle it has to maintain the same angle right? So my question is assuming it hit 150 on 35th day, how many days would it take to reach point A( where the y-coordinate of A is always 0.618 x 200)?

Please explains: "the next 10days (so 35th day from the start (0,0)) it reaches 150 "
Are you saying the rate of increase of 200 in 25 days now becomes the rate of decrease? But 200/25 is 8 per day and in 10 days the decrease will be 80, so y should be 200 - 80 = 120, not 150.
Are any of the angles marked as equal?

 

[meenabala]

Please explains: "the next 10days (so 35th day from the start (0,0)) it reaches 150 "
Are you saying the rate of increase of 200 in 25 days now becomes the rate of decrease? But 200/25 is 8 per day and in 10 days the decrease will be 80, so y should be 200 - 80 = 120, not 150.
Are any of the angles marked as equal?

Dr. Peterson,

I was just giving an example. you can go with what you've drawn on your chart. let's say it reaches 150 in 50 days (counting from the starting point.) can we extrapolate this line to point A and figure out how many days it would reach point A? because the angle at the top remains the same?

 

[Dr.Peterson]

Basically the line from (0,0) to (25,200) gets built first. Assume that we have number of days on the X-axis. so the line from (0,0) to (25,200) get built first in 25 days. Then the line from the top to point A (the y-coordinate of which will always be at 0.618 x 200 ) gets built few points down every subsequent days, until it gets to A. But the initial angle at which that line from (25,200) start out building, maintains that until it gets to point A. So for example, after reaching a top at 200 in 25 days, the next 10days (so 35th day from the start (0,0)) it reaches 150. So the line at 35th day will have a coordinates of (35,150). So for that line to be a side of a triangle it has to maintain the same angle right? So my question is assuming it hit 150 on 35th day, how many days would it take to reach point A( where the y-coordinate of A is always 0.618 x 200)?

So you're adding a new constraint, that line BA has to pass through point D=(35, 150). That makes the problem solvable. Why didn't you mention that before??



Also, you said h₂ = 0.382h, which is 76.4; it's h₁ that's equal to 0.618h = 123.6. Which is correct?

If it's as I drew it, then the SLOPE from B to A has to be the same as the slope from B to D. The latter is (150-200)/(35-25) = -50/10 = -5. The former is (h2-200)/(x-25) = (76.4-200)/(x-25).

Setting this equal to -5, we solve

(76.4-200)/(x-25) = -5​

-123.6/(x-25) = -5​

-123.6 = -5(x-25)​

-123.6 = -5x + 125​

5x = 125+123.6 = 248.6​

x = 49.72​

​

x₂ = x-x₁ = 49.72-25 = 24.72​

 

[meenabala]

Dr. Peterson,

Thank you so much. As I said, I am not good at math and I apologize for not explaining the problem in the first place. This gives me an idea as to how to tackle similar problems. Thank you so much once again

Regards

M Bala

 

[nasi112]

So you're adding a new constraint, that line BA has to pass through point D=(35, 150). That makes the problem solvable. Why didn't you mention that before??

View attachment 27349

Also, you said h₂ = 0.382h, which is 76.4; it's h₁ that's equal to 0.618h = 123.6. Which is correct?

If it's as I drew it, then the SLOPE from B to A has to be the same as the slope from B to D. The latter is (150-200)/(35-25) = -50/10 = -5. The former is (h2-200)/(x-25) = (76.4-200)/(x-25).

Setting this equal to -5, we solve

(76.4-200)/(x-25) = -5​

-123.6/(x-25) = -5​

-123.6 = -5(x-25)​

-123.6 = -5x + 125​

5x = 125+123.6 = 248.6​

x = 49.72​

​

x₂ = x-x₁ = 49.72-25 = 24.72​

but the point [MATH](35,150)[/MATH]made this problem a simple geometry?