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The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Determine the sides of the triangles.

Hrishant Goswami , 10 Years ago
Grade 10
anser 1 Answers
Jitender Pal

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
Let the sides of ∆be n, n + 1, n + 2 where n∈ N.
Let a = n, b = n + 1, c = n + 2
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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 . . . . . . . . . . . . . . . . . . . . . . . . . (1)
In ∆ ABC by Cosine Law, we get
Cos θ = (n + 1)2 + (n + 2)2 – n2/2(n + 1) (n + 2) . . . . . . . . . . . . . . . . . . . (2)
Comparing the values of cos θ from (1) and (2), 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 – n3
⇒ 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.

Thanks
Jitender Pal
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