Question

ABC is a triangle with \(\angle BAC = 60^{\circ}\). A point P lies on one-third of the way from B to C, and AP bisects \(\angle BAC\). \(\angle APC = ?\)

• A. 30°
• B. 45°
• C. 60°
• D. 120°

Hint:

AP bisects the 60 degree angle at A, and P splits BC in the ratio 1:2. Use the sine rule in triangles ABP and ACP to first find the base angles B and C of the triangle, then get angle APC from the angle sum of triangle APC.

Answer 1

The Correct Option is D

Step 1: Set up the given information.
In triangle ABC, the vertex angle is \(\angle BAC = 60^{\circ}\), and AP bisects this angle, so \(\angle BAP = \angle PAC = 30^{\circ}\). Point P lies on BC such that P is one third of the way from B to C, which means \(BP = \frac{1}{3}BC\) and \(PC = \frac{2}{3}BC\), so \(BP:PC = 1:2\).

Step 2: Connect the side ratio to the angle bisector using the ratio lemma.
For a cevian AP drawn from A to a point P on BC, the ratio lemma states \(\dfrac{BP}{PC} = \dfrac{AB}{AC} \cdot \dfrac{\sin \angle BAP}{\sin \angle PAC}\).
Since AP bisects angle A, \(\angle BAP = \angle PAC = 30^{\circ}\), so the two sine terms are equal and cancel out. This leaves \(\dfrac{BP}{PC} = \dfrac{AB}{AC}\), so \(AB:AC = 1:2\).

Step 3: Find the base angles B and C using the Law of Sines.
In triangle ABC, side AB is opposite angle C and side AC is opposite angle B, so by the Law of Sines, \(\dfrac{AB}{AC} = \dfrac{\sin C}{\sin B}\). Using \(AB:AC = 1:2\), this gives \(\sin B = 2\sin C\).
The angles of the triangle add up to 180 degrees, so \(B + C = 180^{\circ} - 60^{\circ} = 120^{\circ}\), which means \(B = 120^{\circ} - C\).
Substituting, \(\sin(120^{\circ} - C) = 2\sin C\). Expanding the left side gives \(\sin120^{\circ}\cos C - \cos120^{\circ}\sin C = 2\sin C\).
Since \(\sin120^{\circ} = \frac{\sqrt3}{2}\) and \(\cos120^{\circ} = -\frac{1}{2}\), this becomes \(\frac{\sqrt3}{2}\cos C + \frac{1}{2}\sin C = 2\sin C\), so \(\frac{\sqrt3}{2}\cos C = \frac{3}{2}\sin C\).
Dividing both sides by \(\frac{3}{2}\cos C\) gives \(\tan C = \frac{1}{\sqrt3}\), so \(C = 30^{\circ}\), and then \(B = 120^{\circ} - 30^{\circ} = 90^{\circ}\).

Step 4: Find angle APC from the angle sum of triangle APC.
Triangle APC has angle \(\angle PAC = 30^{\circ}\) (half of angle A), and angle \(\angle ACP\), which is the same as angle C of the full triangle since P lies on BC, so \(\angle ACP = 30^{\circ}\).
The three angles of triangle APC add up to 180 degrees, so \(\angle APC = 180^{\circ} - 30^{\circ} - 30^{\circ} = 120^{\circ}\).

Final Answer:
The angle APC measures 120 degrees. \[ \boxed{\angle APC = 120^{\circ}} \]