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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The sides of a $\Delta$le are three consecutive natural numbers and its largest angle is twice the smallest one.Determine the sides of the triangle


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Step 1:
Let the sides of $\Delta ABC$ be $x,x+1$ and $x+2$ where $x > 0$ and it is a natural number.
Let $a=x+2,b=x+1,c=x$
Since $x>0$
Therefore $a > b> c\Rightarrow A > B > C$
Let the smallest angle be $\theta$ then $C=\theta$ and greatest angle $A=2\theta$
Applying sine rule in $\Delta$le ABC we get,
$\large\frac{x}{\sin\theta}=\frac{x+2}{\sin 2\theta}=\frac{x+1}{\sin (180^{\large\circ}-3\theta)}$
Step 2:
Let consider first and second terms
$\large\frac{x}{\sin\theta}=\frac{x+2}{\sin 2\theta}$
$\sin 2\theta=2\sin\theta\cos\theta$
$\Rightarrow 2\cos\theta=\large\frac{x+2}{x}$
Step 3:
Let consider first and third terms we get
$\large\frac{x}{\sin\theta}=\frac{x+1}{\sin 3\theta}$
$\Rightarrow \large\frac{2x+1}{x}$
Step 4:
On simplifying we get,
Or $x^2-3x-4=0$
$\Rightarrow x=4$ or $x=-1$
$\Rightarrow x=4$
Hence the sides of $\Delta$le are $4,5,6$
Hence (a) is the correct answer.
answered Oct 21, 2013 by sreemathi.v
edited Jan 10, 2014 by balaji.thirumalai

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