Question

In a ΔABC, the internal bisector of angle A meets BC at D. If AB = 4, AC = 3 and ∠A = 60º, then the length of AD is

(a) 2√3 

(b) \(\frac{12\sqrt3}{7}\)

(c) \(\frac{15\sqrt3}{8}\)

(d) \(\frac{6\sqrt3}{7}\)

Answer 1

(b) \(\frac{12\sqrt3}{7}\)

Let BC = x and AD = y, then as per bisector theorem, 

\(\frac{BD}{DC}= \frac{AB}{AC}\) = \(\frac{4}{3}\)

∴ BD = \(\frac{4x}{7}\) and DC = \(\frac{3x}{7}\)

Now in ΔABD using cosine rule, 

cos 30º = \(\frac{AB^2+AD^2-BD^2}{2\times{AB}\times{AD}}\)

In ΔACD using the cosine rule,

 cos 30º = \(\frac{AC^2+AD^2-CD^2}{2\times{AC}\times{AD}}\)

Also in ΔABC, cos 60º = \(\frac{AB^2+AC^2-BC^2}{2\times{AB}\times{AC}}\)

⇒ \(\frac{1}{2}\) = \(\frac{16+9-BC^2}{2\times4\times3}\) ⇒ 12 = 25 - BC²

⇒ BC² = 13 ⇒ x² = 13

∴ Subtracting eqn (ii) from (i), we get

\(\sqrt3\,y=7-\frac{7x^2}{49}=7-\frac{7\times13}{49}=7-\frac{13}{7}=\frac{49-13}{7}=\frac{36}{7}\)

⇒ y = \(\frac{36}{7\times\sqrt3}\) = \(\frac{12\sqrt3}{7}\).