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: Translate the given ratio into a coordinate setup.
Since P lies one third of the way from B to C, take BC = 3 units so that BP = 1 and PC = 2. Place B at the origin (0, 0) and C at (3, 0), so P sits at (1, 0).

Step 2: Find the base angles using the sine rule in the two smaller triangles.
AP bisects the 60 degree angle at A, so \(\angle BAP = \angle PAC = 30^{\circ}\). Applying the sine rule in triangle ABP, \(BP = \dfrac{AP\sin30^{\circ}}{\sin(\angle ABP)}\), and in triangle ACP, \(PC = \dfrac{AP\sin30^{\circ}}{\sin(\angle ACP)}\).
Since \(\angle ABP\) is just angle B of the triangle and \(\angle ACP\) is just angle C (P lies on BC), dividing the two equations gives \(\dfrac{BP}{PC} = \dfrac{\sin C}{\sin B}\). With \(BP:PC=1:2\), this gives \(\sin B = 2\sin C\).
Combined with \(B + C = 120^{\circ}\), solving this pair of equations the same way as before gives \(C = 30^{\circ}\) and \(B = 90^{\circ}\).

Step 3: Place point A using these angles.
Since angle B is 90 degrees, AB is perpendicular to BC, so A sits directly above B on the y-axis: \(A = (0, h)\).
In the right triangle formed at B, \(\tan(\angle C) = \dfrac{AB}{BC}\), so \(\tan30^{\circ} = \dfrac{h}{3}\), giving \(h = 3\tan30^{\circ} = \sqrt3\). So \(A = (0, \sqrt3)\).

Step 4: Compute angle APC using vectors.
The vector from P to A is \(\vec{PA} = (0 - 1,\ \sqrt3 - 0) = (-1, \sqrt3)\), and the vector from P to C is \(\vec{PC} = (3-1,\ 0-0) = (2, 0)\).
Using the dot product formula, \(\cos(\angle APC) = \dfrac{\vec{PA}\cdot\vec{PC}}{|\vec{PA}||\vec{PC}|} = \dfrac{(-1)(2) + (\sqrt3)(0)}{\sqrt{1+3}\times 2} = \dfrac{-2}{4} = -\dfrac{1}{2}\).

Final Answer:
Since \(\cos(\angle APC) = -\dfrac{1}{2}\), angle APC equals 120 degrees. \[ \boxed{\angle APC = 120^{\circ}} \]