A) 1, 2, 3
B) 2, 3, 4
C) 3, 4, 5
D) 4, 5, 6
Correct Answer: D
Solution :
Let the sides of \[\Delta ABC\]be \[a=n,b=n+1,c=n+2\], where n is a natural number. Then C is the greatest and A the least angle. As given \[C=2A\]. \ \[\sin C=\sin 2A=2\sin A\cos A\] \ \[kc=2ka\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]or \[b{{c}^{2}}=a\text{ }({{b}^{2}}+{{c}^{2}}-{{a}^{2}})\] Substituting the values of a, b, c, we get \[(n+1){{(n+2)}^{2}}=n[{{(n+1)}^{2}}+{{(n+2)}^{2}}-{{n}^{2}}]\] or \[(n+1){{(n+2)}^{2}}=n\text{ }({{n}^{2}}+6n+5)=n(n+1)(n+5)\] Since \[n\ne -1\], we can cancel \[n+1\]. Thus \[{{(n+2)}^{2}}=n\text{ }(n+5)\]or \[{{n}^{2}}+4n+4={{n}^{2}}+5n\] This gives \[n=4,\]Hence the sides are 4, 5 and 6.
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