Let quadrilateral ABCD satisfy ∠ B A C = ∠ C A D = 2 ∠ A C D = 40 ∘ and ∠ A C B = 70 ∘ . Find ∠ A D B.

Question

Let quadrilateral ABCD satisfy   ∠  B  A  C  =  ∠  C  A  D  =  2    ∠  A  C  D  =      40    ∘   and   ∠  A  C  B  =      70    ∘  . Find   ∠  A  D  B.

Dalton Lester · Accepted Answer

Step 1  ≈  77  ,      34          ∘      , soStep 2                    sin        ⁡        80                    sin        ⁡        40                                sin        ⁡        (        70        −        x        )                    sin        ⁡        x                                sin        ⁡        20                    sin        ⁡        90              =  1After some manipulation we get   cot  ⁡  x  =  tan  ⁡  20  +            2              cos        ⁡        10                ⟹    x  =  .  .  .

Joel French · Accepted Answer

Since   ∠  A  C  B  =  ∠  A  B  C  =      70    ∘  , the triangle ABC is isosceles and               A      B        ¯    =            A      C        ¯  .By applying the law of sines to the triangle ACD, we get that:            A      D        ¯    =            A      C        ¯    ⋅                                                  sin          ⁡          ∠          A          C          D                                                                  sin          ⁡          ∠          A          D          C                      =            A      C        ¯    ⋅                                                  sin          ⁡                      20            ∘                                                                            sin          ⁡                      120            ∘                                =                                                  2                                    A              C                        ¯                    sin          ⁡                      20            ∘                                                                                        3                                  .And, by applying the law of sines to the triangle ABD, we get that:            A      D        ¯    sin  ⁡  ∠  A  D  B  =            A      B        ¯    sin  ⁡  ∠  A  B  D    .        (    ∗    )  Let     α  =  ∠  A  D  B    ..Step 2Since               A      D        ¯    =                                                  2                                    A              C                        ¯                    sin          ⁡                      20            ∘                                                                                        3                                ,               A      B        ¯    =            A      C        ¯     and     ∠  A  B  D  =      100    ∘    −  α    ,  , the equality (∗) turns into:                                                  2                                    A              C                        ¯                    sin          ⁡                      20            ∘                    sin          ⁡          α                                                                              3                                =            A      C        ¯    sin  ⁡  (      100    ∘    −  α  )    ,  2  sin  ⁡      20    ∘    sin  ⁡  α  =      3    sin  ⁡  (      90    ∘    +      10    ∘    −  α  )    ,  4  sin  ⁡      10    ∘    cos  ⁡      10    ∘    sin  ⁡  α  =      3    cos  ⁡  (      10    ∘    −  α  )    ,  4  sin  ⁡      10    ∘    cos  ⁡      10    ∘    sin  ⁡  α  =      3        (    cos    ⁡          10      ∘        cos    ⁡    α    +    sin    ⁡          10      ∘        sin    ⁡    α    )      ,  4  sin  ⁡      10    ∘    sin  ⁡  α  =      3        (    cos    ⁡    α    +    tan    ⁡          10      ∘        sin    ⁡    α    )      ,      (    4    sin    ⁡          10      ∘        −          3        tan    ⁡          10      ∘        )    sin  ⁡  α  =      3    cos  ⁡  α    ,  tan  ⁡  α  =                                                              3                                                                            4          sin          ⁡                      10            ∘                    −                      3                    tan          ⁡                      10            ∘                                  .Hence,   ∠  A  D  B  =  α  =  arctan  ⁡      (                                                                                3                                                                                                  4            sin            ⁡                          10              ∘                        −                          3                        tan            ⁡                          10              ∘                                            )    ≃    ≃  77  ,      3361794    ∘    .

Let quadrilateral ABCD satisfy angle BAC= angle CAD=2 angle ACD=40^circ and angle ACB=70^circ. Find angle ADB.

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