Discussion

Explanation:

Applying cosine rule in ∆NPL, we get;
 
Cos60 = 1/2 = [NP² + NL² − PL²]/(2 × NP × NL)
 
∴ 1/2 = [(2x)² + (3x)² − PL²]/(2 × 2x × 3x)
 
∴ PL = x√7.
 
Using angle bisector theorem, MN/NL = MP/PL
 
So, MN/3x = MP/x√7
 
∴ MN = MP × (3/√7) ...(I)
 
Applying cosine rule in ∆MNP, we get;
 
Cos60 = 1/2 = [MN² + NP² − MP²]/(2 × MN × NP)
 
Let MN = a, so using (I) MP = (√7/3) a or MP² = (7/9)a².
 
∴ 1/2 = [a² + 4x²− (7/9)a² ]/(2 × a × 2x)
 
∴ a² − 9ax + 18x² = 0  
 
∴ a = 6x or 3x. 
 
If a = 3x, then MN = NL = 3x, so ∆MNL becomes isosceles and hence ∠NPL = 90°, which means that NL² = NP² + PL².
 
NL² = 9x² and  NP² + PL² = 4x² + (x²/7) = (29/7)x² ≠  9x². Hence a ≠ 3x.
 
∴ a = 6x.
 
NP : NL: MN = 2x : 3x : 6x = 2 : 3 : 6.
 
Hence option (b).
         
Note: We can use the direct formula for the length of the angle bisector L. Consider the triangle drawn below

L = (2abCosQ)/(a + b)

So, 2x = (2 × MN × 3x × Cos60)/(MN + 3x)

Solving, we get; MN = 6x.

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