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Ask iit jee aieee pet cbse icse state board experts Expert Question: Question on Arithmetic progressions(involving trigonometry too!)?
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[Post New]14 Jul 2008 07:28:10 IST
    Subject: Question on Arithmetic progressions(involving trigonometry too!)?
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circumvention (0)

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If the sides of a triangle are in Arithmetic Progression and the greatest angle exceeds the smallest by α, show that the sides are in the ratio 1 - x : 1 : 1 + x , where x = √(1 - cos α)(7 - cos α)?


How to do it? I have no idea whatsoever.
    
[Post New]14 Jul 2008 11:52:52 IST
    Subject: Re:Question on Arithmetic progressions(involving trigonometry too!)?
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allamraju (3415)

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See this.I answered it here.

http://www.goiit.com/posts/list/trignometry-solution-of-triangles-66370.htm#327647
MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC.
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[Post New]14 Jul 2008 13:11:58 IST
    Subject: Re:Question on Arithmetic progressions(involving trigonometry too!)?
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sboosy (3046)

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\mbox{Let the sides of the triangle be} \ x-d,x,x+d \\ \\ \mbox{Dividing by x,ratio of sides becomes} \ 1-\frac{d}{x},1,1+\frac{d}{x} \\ \\ \mbox{Thus we should find the ratio} \ \frac{d}{x} \ \mbox{to obtain final answer} \\ \\ \mbox{Let B be the greatest angle and C the smallest} \\ \\ \Rightarrow B=C+\alpha \\ \\ \mbox{Using} \ A+B+C=\pi \\ \\ A=\pi+\alpha-2B .......(1) \\ \\ C=B-\alpha .......(2) \\ \\ \mbox{As per sine rule} \ \frac{x}{\sin(\pi+\alpha-2B)} = \frac{x+d}{\sin(B)} = \frac{x-d}{\sin(B-\alpha)} \\ \\ \Rightarrow \frac{x}{\sin(2B-\alpha)} = \frac{x+d}{\sin(B)} = \frac{x-d}{\sin(B-\alpha)} .....(3) \\ \\ \\ \mbox{Now using...that if} \ \frac{a}{b}=\frac{c}{d} \Rightarrow \frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d} \\ \\ \\ \Rightarrow \frac{x+d}{\sin(B)} = \frac{2x}{\sin(B)+\sin(B-\alpha)} \\ \\ \\ \Rightarrow \frac{d}{x} = \frac{\sin(B)-\sin(B-\alpha)}{\sin(B)+\sin(B-\alpha)} =\frac{\tan(\frac{\alpha}{2})}{\tan(\frac{2B-\alpha}{2})} = \frac{\tan(\frac{\alpha}{2})}{\tan(\frac{B+C}{2})} ..... (4) \\ \\ \mbox{Using AP} \ 2\sin(B)=\sin(A)+\sin(C) \ \Rightarrow 2\cos(\frac{B+C}{2})=\cos(\frac{\alpha}{2}) .....(5) \\ \\ \frac{d^2}{x^2} = \frac{\tan^2(\frac{\alpha}{2})}{\tan^2(\frac{B+C}{2})} = \frac{\sin^2(\frac{\alpha}{2})}{4-\cos^2(\frac{\alpha}{2})} \ (\mbox{Using 4 and 5}) \\ \\ = \frac{1-\cos(\alpha)}{7-\cos(\alpha)} \\ \\ \Rightarrow \frac{d}{x} = \sqrt\frac{1-\cos(\alpha)}{7-\cos(\alpha)}
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[Post New]14 Jul 2008 13:37:57 IST
    Subject: Re:Question on Arithmetic progressions(involving trigonometry too!)?
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elessar_iitkgp (2220)

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Well answered


Take a salute


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