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In right triangle ABC, we have \(\angle BAC = 90^\circ\) and D is on \(\overline{AC}\) such that \(\overline{BD}\) bisects \(\angle ABC\). If AB = 12 and BC = 15, then what is \(\cos \angle BDC\)?

 Apr 14, 2020
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By the law of cosines, cos BDC = 5/8.

 Apr 14, 2020

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