Question

ABC is a triangle with ∠BAC = 60º. A point P lies on onethird of the way from B to C and AP bisects ∠BAC. ∠APC equals

(a) 90º 

(b) 45º 

(c) 60º 

(d) 120º

Answer 1

(d) 120º

Let BP = x. Then PC = 2x 

∵ AP bisects ∠BAC, 

By the angle bisector theorem,

\(\frac{AB}{AC}=\frac{BP}{PC}\) = \(\frac{1}{2}\)

Using the sine formula, \(\frac{AC}{sin\,B}=\frac{BA}{sin\,C}\)

⇒ \(\frac{sin\,C}{sin\,B}\) = \(\frac{BA}{AC}\) = \(\frac{1}{2}\)

⇒ \(\frac{sin\,C}{sin\,B}\)  = \(\frac{\frac{1}{2}}{1}\) ⇒ sin C = \(\frac{1}{2}\) and sin B = 1

⇒ ∠C = 30° and ∠B = 90° 

∴ In ΔAPC, APC = 180º – (30º + 30º) = 120º