Topic: Calculate Adjacent and opposite of Triangle segments' lengths

[Coder]

Hello experts .

I am trying to Calculate the adjacent and opposite lengths of a Triangle segments , but I found out the result is short in millimeters which is not correct at all .
This is what I wrote but it's not 100% correct , can anyone correct it for me please ?

Code: [Select]

(setq chord 10.)
(setq a (expt chord 2)
      b (/ a 2.)
      c (sqrt b)
)


[Lee Mac]

You are assuming that the triangle has a right-angle... more information is required for a unique solution.

[Coder]

You are assuming that the triangle has a right-angle... more information is required for a unique solution.

Hi Lee ,

Yes it is right angled triangle .
Thanks

[Lee Mac]

You are assuming that the triangle has a right-angle... more information is required for a unique solution.

Yes it is right angled triangle .

If so, then by Pythagoras your solution is correct:

Code - Auto/Visual Lisp: [Select]

1. _$ (setq x 10.0)
2. 10.0
3. _$ (sqrt (* x x 0.5))
4. 7.07107

[Coder]

You are assuming that the triangle has a right-angle... more information is required for a unique solution.

Yes it is right angled triangle .

If so, then by Pythagoras your solution is correct:

Code - Auto/Visual Lisp: [Select]

1. _$ (setq x 10.0)
2. 10.0
3. _$ (sqrt (* x x 0.5))
4. 7.07107

I see it is the same result , but the triangle is not closed correctly  , it still opened

[Lee Mac]

I see it is the same result , but the triangle is not closed correctly  , it still opened

The result shown in the console is only an approximation to 6 significant figures, though the result is still calculated using the maximum precision afforded by a double:

Code - Auto/Visual Lisp: [Select]

1. _$ (rtos (sqrt (* x x 0.5)) 2 16)
2. "7.071067811865475"

However, since the true result is likely irrational, any decimal representation will only ever be an approximation.

[Coder]

I see it is the same result , but the triangle is not closed correctly  , it still opened

The result shown in the console is only an approximation to 6 significant figures, though the result is still calculated using the maximum precision afforded by a double:

Code - Auto/Visual Lisp: [Select]

1. _$ (rtos (sqrt (* x x 0.5)) 2 16)
2. "7.071067811865475"

However, since the true result is likely irrational, any decimal representation will only ever be an approximation.

Nice , that changed the result to be correct now although the decimal number before was with 4 digits ALSO !!! , and this is correct now .

Code: [Select]

(setq chord 10.)
(setq a (expt chord 2)
      b (/ a 2.)
      c (read (rtos (sqrt b) 2 4))
)

That is strange to be corrected by rtos function  , I did not get it why !

[Lee Mac]

Code: [Select]

(setq chord 10.)
(setq a (expt chord 2)
      b (/ a 2.)
      c (read (rtos (sqrt b) 2 4))
)
That is strange to be corrected by rtos function  , I did not get it why !

The result is always calculated using the maximum available precision (in this case ~ 16 dp); the result returned to the command-line or console is truncated for convenience, however, the variable still contains the result expressed to the full precision.

By using the rtos function to express the result as a string to four decimal places, you have rounded the result.

[Coder]

Great Lee ,

Many thanks professor . 

[Lee Mac]

You're welcome