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

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