 |
|
|
|
|
|
||
|
|
||
|
|
||
|
|
|
|
|
|
||
|
|
||
|
|
||
|
|
|
|
|
|||||||||||||||||
Add Forum to Favorites | Send Topic To a Friend | View Forum FAQ | Track this topic |
Last Thread Next Thread |
| isosceles triangle and vertex angle |
|
 |
SpreeTree Member since: 8/9/2003 From: Warwick, United Kingdom |
|||
 Posted - 1/19/2005 4:56:37 AM |
||||
| Hi Guys I have a probably simple problem I need a solution to. I have had a good look on google but to no avail, so any links would be good :) I have an isosceles, and know the length of all three sides, but no angles. I need to know, in radians, the vertex angle (the angle between the two equal sides). Anyone know of any links, or solutions to this. I did have a solution using the right angled triangle on the side of the isosceles triangle, but this doesnt seem to work correctly. Thanks Spree |
||||
|
||||
|
blizzard999 Member since: 4/21/2003 From: Italy |
|||
|
Posted - 1/19/2005 6:04:09 AM |
||||
Quote: Let be l the lenght of the equal edges b the third edge (the base) a the angle between the two equal edges. you know that
Sorry for my poor ASCII art!
***
** *
l ** * b/2
** *
**a/2-----*
** *
**
...
l * sin(a/2) = b/2
Then
a = 2*asin(b/(2*l))
And I hope this is not your homework ;) [Edited by - blizzard999 on January 19, 2005 5:04:31 AM] |
||||
|
||||
|
SpreeTree Member since: 8/9/2003 From: Warwick, United Kingdom |
|||
|
Posted - 1/19/2005 7:22:26 AM |
||||
| Thanks a lot. Of course now I see what u have done, it makes it pretty obvious, and me pretty simple ;) And no, its not my homework. I think if it was, I would be the oldest kid in the class ;) Its actually for computing the length of an orbital body so the base of a circle is the same size, regadless of the distance from the centre object... Thanks Spree |
||||
|
||||
| All times are ET (US) |
Last Thread Next Thread |
|
