All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Cot Ratio Conclusion (Posted on 2014-10-18) Difficulty: 3 of 5
BC, CA and AB represents three sides of a triangle having lengths a, b and c respectively. Angles A, B and C are opposite respectively to the sides BC, CA and AB

Given that: a2 + b2= 2015* c2

Find the value of :
   cot  C
-------------
cot A + cot B

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 2 of 2 |
a2 + b2 = 2015c2 and the rule c2 = a2 + b2 – 2ab*cos C,

together give:  ab*cos C =1007c2.                                   (1)

     cot C                      sin A * sin B * cot C
---------------      =    ----------------------------------
cot A + cot B            sin B * cos A  +  sin A * cos B

                                sin A * sin B * cos C
                        =    -------------------------- 
                                  sin C * sin(A + B)

                                sin A * sin B * cos C
                        =    --------------------------               (2)
                                         sin2C

Now, since   a/sin A = b/sin B = c/sin C, (2) becomes

     cot C
---------------      =   ab*cos C / c2  =  1007       using (1)
cot A + cot B    


  Posted by Harry on 2014-10-22 18:05:57
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information