Question

The sides of a triangle are consecutive integers n, n +1 and n + 2 and the largest angle is twice the smallest angle. Find n.

Answer 1

Let the sides of ∆be n, n + 1, n + 2 where n ∈ N. 

Let a = n, b = n + 1, c = n + 2

Let the smallest, angle ∠A = θ then the greatest ∠C = 2θ. 

In ∆ABC by applying Sine Law we get, 

Sin θ/n = sin 2 θ/n + 2 

⇒ sin θ/n = 2 sin θ cos θ/ n + 2 

⇒ 1/n = 2 cos θ/n + 2 (as sin θ ≠ 0) 

⇒ cos θ = n + 2/2n ... (i) 

In ∆ ABC by Cosine Law, we get 

Cos θ = (n + 1)² + (n + 2)² – n²/2(n + 1) (n + 2) ... (ii) 

Comparing the values of cos θ from (i) and (ii), we get 

(n + 1)² + (n + 2)² – n²/2(n + 1) (n + 2) = n + 2/2n 

⇒ (n + 2)² (n + 1) = n(n + 2)² + n(n + 1)² – n² 

⇒ n (n + 2)² (n + 2)² = n(n + 2)² + n(n + 1)² – n³ 

⇒ n² + 4n + 4 = n³ + 2n² + n – n³ 

⇒ n² – 3n – 4 = 0 ⇒ (n + 1) (n – 4) = 0 

⇒ n = 4 (as n ≠ - 1)

∴ Sides of ∆ are 4, 4 + 1, 4 + 2, i.e. 4, 5, 6.