Category: Trigonometry

Question Number 209560 by MM42 last updated on 15/Jul/24 $${In}\:{the}\:{triangle}\:{ABC}\:;\:{cos}\left({B}−{C}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${Show}\:{that}\::\:\:\frac{\mathrm{1}−\mathrm{3}{cos}\left({B}+{C}\right)}{\mathrm{6}{sinBcosC}}={tanC} \\ $$$$ \\ $$ Answered by Spillover last updated on 14/Jul/24 $$\:\:\frac{\mathrm{1}−\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{6}{sinBcosC}}={tanC} \\…

Question Number 209458 by SonGoku last updated on 10/Jul/24 Commented by SonGoku last updated on 10/Jul/24 $$\mathrm{l}\:=\:?\: \\ $$$$\mathrm{BC}\:=\:\:\mathrm{l}\:+\:\mathrm{200m} \\ $$$$\mathrm{B}\hat {\mathrm{A}C}\:=\:\mathrm{120}° \\ $$$$\mathrm{A}\hat {\mathrm{B}C}\:=\:\mathrm{38}°\mathrm{12}'\mathrm{48}''…

Question Number 209211 by SonGoku last updated on 03/Jul/24 Commented by A5T last updated on 03/Jul/24 $${AC}=\sqrt{{AB}^{\mathrm{2}} +{BC}^{\mathrm{2}} −\mathrm{2}{AB}×{BCcos}\mathrm{123}}\approx\mathrm{74}.\mathrm{738} \\ $$ Terms of Service Privacy…

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Question Number 208755 by efronzo1 last updated on 22/Jun/24 Answered by mr W last updated on 22/Jun/24 Commented by mr W last updated on 22/Jun/24…