In Δ A B C,such 3 sin 2 ⁡ B + 7 sin 2 ⁡ C = 2 sin 2 ⁡ A + 2 sin ⁡ A sin ⁡ B sin ⁡ C , and such a = | B C | = 1, show that sin ⁡ A = 5 5 I think this problem 3 b 2 + 7 c 2 = 2 a 2 + 2 sin ⁡ A b c = 2 + sin ⁡ A ⋅ b cwhere | A C | = b , | A B | = c

Question

In   Δ  A  B  C,such   3      sin    2    ⁡      B    +  7      sin    2    ⁡      C    =  2      sin    2    ⁡      A    +  2  sin  ⁡      A    sin  ⁡      B    sin  ⁡      C  , and such   a  =      |    B  C      |    =  1, show that  sin  ⁡      A    =                    5            5      I think this problem  3      b    2    +  7      c    2    =  2      a    2    +  2  sin  ⁡      A    b  c  =  2  +  sin  ⁡      A    ⋅  b  cwhere       |    A  C      |    =  b  ,      |    A  B      |    =  c

Pedro Hurley · Accepted Answer

3      b    2    +  7      c    2    =  2      a    2    +  2  sin  ⁡      A    b  c  =  2  +  sin  ⁡      A    ⋅  b  cThis should be  3      b    2    +  7      c    2    =  2      a    2    +  2  sin  ⁡      A    b  c  =  2  +      2    sin  ⁡      A    ⋅  b  cfrom which we have                    (1)                    sin        ⁡        A        =                              3                          b              2                        +            7                          c              2                        −            2                                2            b            c                              By the way, by the law of cosines,                    (2)                    cos        ⁡        A        =                                            b              2                        +                          c              2                        −            1                                2            b            c                              Using that       sin    2    ⁡  A  +      cos    2    ⁡  A  =  1 with (1)(2),            (                        3                      b            2                    +          7                      c            2                    −          2                          2          b          c                    )        2    +            (                                    b            2                    +                      c            2                    −          1                          2          b          c                    )        2    =  1Multiplying the both sides by   (  2  b  c      )    2   gives  (  3      b    2    +  7      c    2    −  2      )    2    +  (      b    2    +      c    2    −  1      )    2    −  (  2  b  c      )    2    =  0which can be written as  10            (              b        2            +                        20                      c            2                    −          7                10            )        2    +      1    10    (  10      c    2    −  1      )    2    =  0from which we have      b    2    +            20              c        2            −      7        10    =  10      c    2    −  1  =  0  ,i.e.  b  =      1          2        ,    c  =      1          10      The claim follows from (1)

In <mi mathvariant="normal">Δ A B C ,such 3 sin 2 </msup>

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