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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: SOT chall. 1
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[Post New]12 Mar 2008 01:09:34 IST
    Subject: Re:SOT chall. 1
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raulrag009 (1205)

Blazing goIITian

Olaaa!! Perrrfect answer. 203  [298 rates]
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My\,attempt\\\\ Let\,a-d\:,a\:,a+d\:be\,the\,sides\,of\,the\,triangle\\\\ a(a-d)\sin{A}\,=\,a(a+d)\sin{B}\\\\ \frac{a-d}{a+d}\,=\,\frac{\sin{B}}{\sin{A}}\\\\ \frac{1-\frac{d}{a}}{1+\frac{d}{a}}\,=\,\frac{\sin{B}}{\sin{(B+\alpha)}}\\\\ Apply Componendo and dividendo\\\\ \frac{2}{\frac{-2d}{a}}\,=\,\frac{\sin{B}+\sin{(B+\alpha)}}{\sin{B}-\sin{(B+\alpha)}}\\\\ hence\\ \frac{d}{a}\,=\,\frac{\sin{(B+\alpha)}-\sin{B}}{\sin{B}+\sin{(B+\alpha)}}\\\\ \frac{d}{a}\,=\,\tan{\frac{\alpha}{2}}\tan{\frac{C}{2}}\\\\ Ratio\;is\\ 1-tan{\frac{\alpha}{2}}:1:1+tan{\frac{C}{2}}
 this reply: 5 points  (with 1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
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