geometry!
posted by please help me out!!!! .
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?

I placed D outside the triangle topright 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°