1 a + b + 1 a + c = 3 a + b + c in a triangleFind the angle α of a triangle with sides a, b and c for which the equality

Question

1          a      +      b        +      1          a      +      c        =      3          a      +      b      +      c       in a triangleFind the angle   α of a triangle with sides a, b and c for which the equality

Valeria Wolfe · Accepted Answer

Step 1  cos  ⁡  α  =                              b          2                +                  c          2                −                  a          2                            2        b        c            after simplifying the given equality and plug in something, but this does not seem to be easy.            1              a        +        b              +            1              a        +        c              =            3              a        +        b        +        c                                  2        a        +        b        +        c                    (        a        +        b        )        (        a        +        c        )              =            3              a        +        b        +        c                (  2  a  +  b  +  c  )  (  a  +  b  +  c  )  =  3  (  a  +  b  )  (  a  +  c  )Step 2The last line, you get       b    2    +      c    2    =      a    2    +  b  c , so the   cos  ⁡  α  =  1      /    2cosα=1/2 .

Logan Wyatt · Accepted Answer

Step 1Multiplying both sides of the equation by  (  a  +  b  )  (  a  +  c  )  (  a  +  b  +  c  ) , we get                     (        a        +        c        )        (        a        +        b        +        c        )        +        (        a        +        b        )        (        a        +        b        +        c        )                            =        3        (        a        +        c        )        (        a        +        b        )                                      a          2                +        a        b        +        2        a        c        +        b        c        +                  c          2                +                  a          2                +        2        a        b        +        a        c        +                  b          2                +        b        c                            =        3                  a          2                +        3        a        b        +        3        a        c        +        3        b        c                            2                  a          2                +        3        a        b        +        3        a        c        +                  b          2                +        2        b        c        +                  c          2                                    =        3                  a          2                +        3        a        b        +        3        a        c        +        3        b        c                            b        c                            =        −                  a          2                +                  b          2                +                  c          2                    Hence, we have   cos  ⁡  α  =      1    2   . Since we know   0  &amp;lt;  α  &amp;lt;  π, we have   α  =                                          π            3

1 a + b </mrow> </mfrac> + 1 a

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