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Question Number 146878 by gsk2684 last updated on 16/Jul/21

in ΔABC if sin^2 A sin B sin C+  cos Bcos C=1 then the triangle is

$${in}\:\Delta{ABC}\:{if}\:\mathrm{sin}\:^{\mathrm{2}} {A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}+ \\ $$$$\mathrm{cos}\:{B}\mathrm{cos}\:{C}=\mathrm{1}\:{then}\:{the}\:{triangle}\:{is} \\ $$

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