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ABC is an isosceles triangle with AB=AC, ∠BAC=96∘. D is a point such that ∠DCA=48∘, AD=BC and angle DAC is obtuse. What is the measure (in degrees) of ∠DAC?

  • geometry! -

    I placed D outside the triangle top-right of A, joined CD and AD
    In triangle ABC , it is easy to see that angle ACB = 42°
    and if angle DCA= 48° , then angle BCD = 90°
    (this helps to make a reasonable sketch)

    Suppose we let AC = 5 (anything will do)
    then by the sine law:
    BC/sin96 = 5/sin42
    BC = 7.43145 (I stored that in my calculator)

    now switch to triangle ACD
    sinD/AC = sin48/AD , but AD = BC = 7.431...
    and AC = 5

    SinD/5 = sin48/7.431..
    sinD = .5
    D = 30° , ahhh!!!

    then angle DAC = 180+48+30 = 102°

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