Members Login
Username 
 
Password 
    Remember Me  
Post Info TOPIC: Corrections for acute/obtuse triangles


Administrator

Status: Offline
Posts: 169
Date:
Corrections for acute/obtuse triangles
Permalink   


To all:


I must apologize for a previous post where I was speaking about finding values to make a triangle into an obtuse or acute triangle.


If given 2 sides of a triangle and asked to find a 3rd side such that it is the smallest/largest value to make an acute/obtuse triangle then do the following:


For an obtuse triangle:


For obtuse triangles, finding the smallest/largest value of the 3rd side is easy.  What you do is find the smallest/largest integer value that makes a triangle.  In other words for the smallest value: |a-b|+1, and for the largest value: (a+b)-1.  This will always be an obtuse triangle.


Ex [1]  Find the largest integral value of x, such that 3,7, and x form an obtuse triangle.



  • Find 7+3-1 = 9.

  • The answer is 9.

  • To find the smallest integral value of x, use 7-3+1 = 5.

For an acute triangle:


To find the values of the third side that would make the triangle an acute triangle you can use the following:


sqrt(a^2-b^2)<x<sqrt(a^2+b^2)


So, if you are looking for the smallest integer value of x such that it forms an acute triangle use sqrt(a^2-b^2) (where a is the largest value) and find the highest integer greater than this square root.  If you are looking for the largest value of x, then use sqrt(a^2+b^2) and find the highest integer less than this square root.


Ex [2]  Find the smallest integral value of x such that 5, x, 9 forms an acute triangle.




  • Since we are looking for the smallest value we use sqrt(a^2-b^2) = sqrt (9^2-5^2) = sqrt (56).  The highest integer value greater than this is 8.  The answer is 8.


  • If we were looking for the largest value, we use sqrt (a^2+b^2) = sqrt (9^2+5^2) = sqrt (106).  The highest integer value less than this is 10.  The answer is 10.

Sorry about the mix up before.  This is the correct way!


Webmaster



__________________
Webmaster


Administrator

Status: Offline
Posts: 169
Date:
Permalink   

Please notice that I changed my original post.  Now the post reads the correct way and I will put this up on my website soon.  Currently I am working on a trig section.  After this I will work on a geometry section which will include this method.


Webmaster



__________________
Webmaster


Senior Member

Status: Offline
Posts: 138
Date:
Permalink   

Thanks...i see...this is a whole lot clearer than the message you posted before. Looking forward to those new sections...hopefully they'll be up before regional


__________________
Page 1 of 1  sorted by
 
Quick Reply

Please log in to post quick replies.

Tweet this page Post to Digg Post to Del.icio.us


Create your own FREE Forum
Report Abuse
Powered by ActiveBoard