Question

please please please answer this question fast.....don't spam or I will report ​

Answer 1

Answer:

Step-by-step explanation:

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

Answer 2

Answer:

plzzz give me brainliest ans and plzzzz follow me