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

Answer: Option B

Let BC = x and Ad = y, then as per bisector theorem,
$$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}=\frac{4}{3}$$
Hence, BD = $$\frac{4\text{x}}{7}$$ and DC = $$\frac{3\text{x}}{7}$$
Now, in ΔABD using cosine rule,
$$\cos 30^{\circ }=\frac{4^2+{\text{y}}^2-\frac{16{\text{x}}^2}{49}}{2\times 3\times \text{y}}$$
$$\text{or,}4\sqrt{3\text{y}}=16+{\text{y}}^2-\frac{16{\text{x}}^2}{49}$$       - - - - - - (i)
Similarly in ΔADC,
$$\cos 30^{\circ }=\frac{3^2+{\text{y}}^2-\frac{9{\text{x}}^2}{49}}{2\times 3\times \text{y}}$$
$$\text{or,}3\sqrt{3\text{y}}=9+{\text{y}}^2-\frac{9{\text{x}}^2}{49}$$       - - - - - - - - (ii)
From equation (i) and (ii), we get
$$\text{y}=\frac{12\sqrt{3}}{7}$$