#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

Arun
25763 Points
4 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
2 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
one year ago
Dear student,
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