Question

In a triangle JKL, the internal bisector of the angle J meets JL = 3 and $\angle J={60}^{\circ }.$ The length JM 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 KL = x and JM = y, Using bisector theorem,

$\frac{KM}{ML}=\frac{JK}{JL}=\frac{4}{3}Thus,KM=\frac{4x}{7}andML=\frac{3x}{7}$

Now, in triangle JKM, using cosine rule,

$Cos{30}^{\circ }=\frac{\{4^2+y^2-\frac{16x^2}{49}\}}{2\times 2y}\Rightarrow 4\sqrt{3}y=\{16+y^2-\frac{16x^2}{49}\}....\left(1\right)$

Similarly in triangle JML,

$Cos{30}^{\circ }=\frac{\{3^2+y^2-\frac{9x^2}{49}\}}{2\times 3y}\Rightarrow 3\sqrt{3}y=\{9+y^2-\frac{9x^2}{49}\}eq\left(2\right)$

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

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

Thus, Option (b) is correct.