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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: SOT challenge 2
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[Post New]9 Mar 2008 12:05:21 IST
    Subject: Re:SOT challenge 2
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hsbhatt (4420)

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Olaaa!! Perrrfect answer. 832  [962 rates]
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\text {The angles are} \ 60^\circ - d, \ 60^\circ, \ 60^\circ + d \ \text {with d positive} \\ \\ 3-4sinA sinC \ = \ 3 -4\sin(60^\circ - d) \sin(60^\circ + d) \\ \\ = 3 - 4(\sin^2(60^\circ) - \sin^2(d)) \ \text {Using the identity} \ \sin(A+B) sin (A-B) = \sin^2A - \sin^2B \\ \\ = 4 \sin^2d \\ \\ \text {Hence} \ \sqrt{3-4sinAsinC} = 2 sind \\ \\ \text {Also} \ |A-C| = 2d \\ \\ \text {Hence the required limit is} \ \lim_{d\rightarrow0} \frac {2sind} {2d} = 2
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