Question

In a triangle $ABC$ the internal bisector of the angle $A$ meets $BC$ at $D$ if $AB=4,AC=3$ and $\angle A={60}^{\circ }$, then the length of $AD$ is

• A. $2\sqrt{3}$
• B. $\frac{12\sqrt{3}}{7}$
• C. $\frac{15\sqrt{3}}{8}$
• D. None of these

Answer 1

The correct option is B $\frac{12\sqrt{3}}{7}$

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

$\frac{BD}{DC}=\frac{AB}{AC}=\frac{4}{3}$

Hence,

$\frac{BD}{DC}=\frac{4x}{7},DC=\frac{3x}{7}$

Now in triangle ABD, using cosine rule,

$\cos {30}^0=\frac{\{4^2+y^2-\left(\frac{16x^2}{49}\right)\}}{2\times 3y}$

$\Rightarrow 4\sqrt{3}y=\{16+y^2-\left(\frac{16x^2}{49}\right)\}$ .....(eqn 1)

Similarly in triangle $ADC$,

$\cos {30}^0=\frac{\{3^2+y^2-\left(\frac{9x^2}{49}\right)\}}{2\times 3y}$

$\Rightarrow 3\sqrt{3}y=\{9^2+y^2-\left(\frac{9x^2}{49}\right)\}$ .....(eqn 2)

From eqn (1) and (2) we get,

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

Hence option B is correct.

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