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 :

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
$${\text{y}} = \frac{{12\sqrt 3 }}{7}$$