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isosceles triangle and vertex angle

Started by SpreeTree Jan 19, 2005 at 3:56 AM 1 replies 1.2k views

Original Post

Jan 19, 2005 at 3:56 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

Jan 19, 2005 at 5:04 AM

Quote:
Original post by SpreeTree
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).

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

Jan 19, 2005 at 6:22 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

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