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13

Part of 1994 AMC 12/AHSME

Problems(1)

1994 AMC 12 #13

Source:

12/30/2011
In triangle ABCABC, AB=ACAB=AC. If there is a point PP strictly between AA and BB such that AP=PC=CBAP=PC=CB, then ∠A=\angle A = [asy] draw((0,0)--(8,0)--(4,12)--cycle); draw((8,0)--(1.6,4.8)); label("A", (4,12), N); label("B", (0,0), W); label("C", (8,0), E); label("P", (1.6,4.8), NW); dot((0,0)); dot((4,12)); dot((8,0)); dot((1.6,4.8)); [/asy] <spanclass=′latex−bold′>(A)</span> 30∘<spanclass=′latex−bold′>(B)</span> 36∘<spanclass=′latex−bold′>(C)</span> 48∘<spanclass=′latex−bold′>(D)</span> 60∘<spanclass=′latex−bold′>(E)</span> 72∘ <span class='latex-bold'>(A)</span>\ 30^{\circ} \qquad<span class='latex-bold'>(B)</span>\ 36^{\circ} \qquad<span class='latex-bold'>(C)</span>\ 48^{\circ} \qquad<span class='latex-bold'>(D)</span>\ 60^{\circ} \qquad<span class='latex-bold'>(E)</span>\ 72^{\circ}
AMC