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In a triangle ABC, if angle C=60 Then prove that 1/(a+c) + 1/(b+c) = 3/(a+b+c) Where the symbols have their usual meaning representing sides of the triangle

In a triangle ABC, if angle C=60
Then prove that 1/(a+c) + 1/(b+c) = 3/(a+b+c)
Where the symbols have their usual meaning representing sides of the triangle

Grade:11

3 Answers

Arun
25750 Points
6 years ago
Try this formula to prove reverse 

a^2 = b^2 + c^2 − 2bc Cos A 
b^2 = c^2 + a^2 − 2ca Cos B 
c^2 = a^2 + b^2 − 2ab Cos C 

u know the angle of C = 60 

So c^2 = a^2 + b^2 − 2ab cosC 
=> c^2 = a^2 + b^2 − 2ab cos 60 => c^2 = a^2 + b^2 − ab 

Simplify 

1/(a+c) + 1/(b+c) = 3 / ( a+b+c ) 

=> (a+b+2c) / (a+c) (b+c) = 3 / (a+b+c) 

=> (a+b+2c) (a+b+c) = 3(a+c) (b+c) 

=> a^2 + b^2 + 2ab + 2c^2 + 3ac + 3bc = 3ab + 3ac + 3bc + 3c^2 

=> a^2 + b^2 + 2ab - 3ab = 3 c^2 - 2 c^2 (cancel 3ac & 3bc) 


=> a^2 + b^2 – ab = c^2 

so both matched hence the proof
Shreyansh
13 Points
4 years ago
 

1/(a+c) + 1/(b+c) = 3 / ( a+b+c ) 

=> (a+b+2c) / (a+c) (b+c) = 3 / (a+b+c) 

=> (a+b+2c) (a+b+c) = 3(a+c) (b+c) 

=> a^2 + b^2 + 2ab + 2c^2 + 3ac + 3bc = 3ab + 3ac + 3bc + 3c^2 

=> a^2 + b^2 + 2ab - 3ab = 3 c^2 - 2 c^2 (cancel 3ac & 3bc) 
Now a^2+b^2-ab-c^2=0
a^2+b^2-c^2=ab
a^2+b^2-c^2/ab=0 
Divided both side by ½
a^2+b^2-c^2/2ab=1/2
CosC=1/2
Cos C=cos 60
C=60   H. P


=> a^2 + b^2 – ab = c^2 
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached solution to your problem.
a2 = b2 + c2 − 2bc Cos A 
b2 = c2 + a2 − 2ca Cos B 
c2 = a2 + b2 − 2ab Cos C 
angle C = 60 
So c2 = a2 + b2 − 2ab cosC 
=> c2 = a2 + b2 − 2ab cos 60
=> c2 = a2 + b2 − ab 
Simplify 
1/(a+c) + 1/(b+c) = 3 / ( a+b+c ) 
=> (a+b+2c) / (a+c) (b+c) = 3 / (a+b+c) 
=> (a+b+2c) (a+b+c) = 3(a+c) (b+c)
=> a2 + b2 + 2ab + 2c2 + 3ac + 3bc = 3ab + 3ac + 3bc + 3c2 
=> a2 + b2 + 2ab - 3ab = 3 c2 - 2 c2 (cancel 3ac & 3bc)
=> a2 + b2 – ab = c2
 
Hope it helps.
Thanks and regards,
Kushagra

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