If a triangle abc (1+a/b+c/b)(1+b/c-c/b)=3 then the angle a is equal to
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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
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