Posted by please help me out!!!! on Monday, February 18, 2013 at 12:36am.
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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