如何计算三角的TAN

Andrew1979 · 发表于 2022-7-5 18:14:57

我搜索了论坛，试图找出这个问题。我怎样锻炼谭？
为什么AutoLISP甚至没有这个功能？

我曾尝试阅读过各种帖子，但都没有任何意义，很惊讶在这个论坛的搜索中没有找到一条帖子

我想做的就是算出三角形的另一边。

我有相邻的长度和角度，但有这么多弧度和度数以及其他垃圾，我就是无法让它工作。

当我使用计算器时，我会在一秒钟内解决它。但不能使用AutoLISP。

[img]https://www.cadtutor.net/forum/image/png;base64，Ivborw0kgoaaaansuheugaaacuaaaaekcaiaadshdmyaafnuleqvr4no2dz47usjfferwvquwehjcathqre9ysglrsecibtxal5faqmgk1vmaskupzpj5nqn5gvkjgjbzepvz08/hwqdhex4jribsnxmekruh6zpicdp4ybf+Hb00/3slqj7fvy0rupnhukkg2i5diqk2o7mbinubtudnzlgouiwzyfghgkisdgaujmodiu/j07bbbjliqko3a0gqzqqzqqzqo4 Qpetp2icaswqjidmbpggwg8pq8hteeiksbgq7sdrbbhb5sp6ohskirbycsh1ymiadidwlt8coeuqicwhzdixnkafenpkny4ciipggwghlibiacjpydoxqwsryejatgnLE2QAkafk6dghgkhkISDbgaUJMoDI0548vfvgUVG79/axpusx334vlsdbeuhkisdb0dknko3ajqbmzywuchaofju5kkngg1jnqsulmzfgriu1a75klmbzuawkkkxk6doggelkiyzgayimipkup07rbcjlarko7a0qqyqeqejh0iieqwaridwjoga4ju/hdhfeigsb2q4stzabr6sp2ohckrybybx6rmez7neege1doit8ntsumekxw6/w48+n0c9q2csszzzz 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eldon · 发表于 2022-7-5 18:30:40

I tried to Google your problem, and the first one was this site.

I am sure there are many more sources of information.

P.S. if you are using whole angles, the hypotenuse length is 662.03

Andrew1979 · 发表于 2022-7-5 18:36:25

thanks, yeah i saw that. Just can't get any of the code to work the way i want it to. I know the info is there, I just don't understand why my code doesn't work

rlx · 发表于 2022-7-5 18:41:47

See this one : https://forums.autodesk.com/t5/visual-lisp-autolisp-and-general/sin-cos-tan/td-p/868027

(btw radians instead of degrees)

gr. Rlx

Lee Mac · 发表于 2022-7-5 18:52:27

Using my Trigonometric Functions:

copy

_$ (* 600.0 (tan (* pi (/ 25.0 180.0))))279.785

Andrew1979 · 发表于 2022-7-5 18:58:51

Ok, I think I have it sorted out. I don't understand why this works exactly or even understand the formula but this is how to work out the opposite side of a right angled triangle when you know the adjacent side and angle

copy

[color=red]([/color][color=royalblue]setq[/color] AngInDegrees[color=seagreen] 25[/color][color=red])[/color][color=gray]; In this example, I set the angle to 25 degrees (not radians)[/color][color=red]([/color][color=royalblue]setq[/color] SIDE [color=seagreen]600[/color][color=red])[/color][color=gray]; In this example I set the opposite side to 600mm)[/color][color=red]([/color][color=royalblue]setq[/color] AngInDegrees-90 [color=red]([/color][color=royalblue]-[/color] AngInDegrees [color=seagreen]90[/color][color=red]))[/color][color=gray];I then minus 90 from the the 25 [/color] [color=red]([/color][color=royalblue]setq[/color] c [color=red]([/color][color=royalblue]/[/color] SIDE [color=red]([/color][color=royalblue]sin[/color] [color=red]([/color][color=royalblue]*[/color] AngInDegrees-90 [color=red]([/color][color=royalblue]/ pi [/color][color=seagreen]180[/color][color=red])))))[/color][color=gray];Do some formula[/color] [color=red]([/color][color=royalblue]setq[/color] b [color=red]([/color][color=royalblue]sqrt [/color][color=red]([/color][color=royalblue]-[/color] [color=red]([/color][color=royalblue]*[/color] c c[color=red])[/color] [color=red]([/color][color=royalblue]*[/color] SIDE SIDE[color=red]))))[/color][color=gray];Do some formula[/color]

What you end up with is a triangle where the opposite side equals 279.78mm. The value of SETQ b = 279.78mm

So it works and if you need to work out this type of triangle this formula will work. Just replace the variable names to suit your own needs.

I wrote all that out so if someone searches for it i the future it may be helpful....and someone might even find it from a google search

Andrew1979 · 发表于 2022-7-5 19:07:03

thanks Lee, you posted as I was writing my last response. i will have to look at that. Than ks everyone.

Lee Mac · 发表于 2022-7-5 19:18:31

Your code is essentially calculating the length of the hypotenuse using the known adjacent side & angle, and then using Pythagoras' Theorem to calculate the remaining unknown side length.

You can omit the subtraction of 90 degrees by simply replacing sin with cos.

[编程交流] 如何计算三角的TAN

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