Given: triangles abc and dbc are isosceles, m∠bdc = 30°, and m∠abd = 155°. find m∠abc, m∠bac, and m∠dbc.

mâ ABC = 35° mâ BAC = 72.5° mâ DBC = 120° Since the diagram wasn't provided, I am making the following assumption: â DBC and â ABC are the vertex angles of the two isosceles triangles. Since, mâ BDC = 30° and triangle DBC is isosceles, then mâ BCD = 30° as well, and mâ DBC = 180° - 30° - 30° = 120° Since mâ ABD = 155°, and mâ ABD = mâ DBC + mâ ABC, that means that mâ ABC = â ABD - â DBC = 155° - 120° = 35° Finally, since mâ ABC = 35° and triangle ABC is isosceles, then mâ BAC = (180° - 35°)/2 = (145°)/2 = 72.5° Pay attention to the assumption in the above calculations. There is a total of 9 different possibilities for this question depending upon what angles are the vertex angles of the triangles. (Actually, only 7 of the 9 potential vertex angles work. 2 of the 9 have the base angles exceed 90 degrees which is impossible for a triangle). Vertex angles â ABC & â DBC: mâ ABC = 35°, mâ BAC = 72.5°, mâ DBC = 120° â ABC & â BDC: mâ ABC = 80°, mâ BAC = 50°, mâ DBC = 75° â ABC & â DCB: mâ ABC = 125°, mâ BAC = 27.5°, mâ DBC = 30° â ACB & â DBC: mâ ABC = 35°, mâ BAC = 35°, mâ DBC = 120° â ACB & â BDC: mâ ABC = 80°, mâ BAC = 80°, mâ DBC = 75° â ACB & â DCB: Impossible â BAC & â DBC: mâ ABC = 35°, mâ BAC = 110°, mâ DBC = 120° â BAC & â BDC: mâ ABC = 80°, mâ BAC = 20°, mâ DBC = 75° â BAC & â DCB: Impossible

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