If 1 a + b + 1 b + c = 3 a + b + c , prove that ∠ B = 60 ∘ My Attempt 1 a + b + 1 b + c = 3 a + b + c a + 2 b + c ( a + b ) ( b + c ) = 3 a + b + c a 2 − a c − b 2 + c 2 = 0How to prove further?

Question

If             1              a        +        b              +            1              b        +        c              =            3              a        +        b        +        c            , prove that   ∠  B  =      60    ∘  My Attempt            1              a        +        b              +            1              b        +        c              =            3              a        +        b        +        c                                a        +        2        b        +        c                    (        a        +        b        )        (        b        +        c        )              =            3              a        +        b        +        c                  a    2    −  a  c  −      b    2    +      c    2    =  0How to prove further?

g2joey15 · Accepted Answer

By the cosine rule,   cos  ⁡  B  =      1    2   iff       b    2    =      a    2    +      c    2    −  a  c. That&#039;s the result you need to try to obtain by rearrangement.

Arraryeldergox2 · Accepted Answer

from your condition we get      a    2    +      c    2    −  a  c  =      b    2  apllying the theorem of cosines we get      a    2    +      c    2    −  a  c  =      a    2    +      c    2    −  2  a  c  cos  ⁡  (  β  )thus we get  cos  ⁡  (  β  )  =      1    2

If <mstyle displaystyle="true" scriptlevel="0"> 1 a + b

Answered question

Answer & Explanation

New Questions in Precalculus