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December 19, 2014

December 19, 2014

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

- geometry! -
**Reiny**, Monday, February 18, 2013 at 12:57amI 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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