
In triangle
ABC,
DB and
DC are angle bisectors and the angle
BAC = 60°. If angle
DCB is 40°, what is the degree measure of angle
BDC? (
Note: Figure not drawn to scale.)
A. 130°
B. 120°
C. 100°
D. 80°
E. 75°
(B) First of all remember that the sum of all angles in a triangle is equal to 180°. We are told that angle
BAC = 60°. That leaves 120° to be split between angles
ABC and
ACB. We know that
DC and
DB are angle bisectors (an angle bisector can be described as a line segment that divides an angle into two equal angles), so the sum of angles DBC and DCB is 120°/2 = 60°. BDC is the third angle in triangle BDC. Therefore it equals 180° – 60° = 120°. The correct answer is B.
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I do not understand why the answer is 120° and not 100°. If as you say, DB and DC are angle bisectors and DCB is 40° degrees (as stated in the question) CBD should also be 40° (which lets CBA and ACB as 60 degrees angles). If all the angles must be 180 degrees 180° – 80° = 100°.