Question Number 209880 by mnjuly1970 last updated on 24/Jul/24 Answered by mahdipoor last updated on 25/Jul/24 $${get}\::\:\:\:{a}={cte}\:\:{and}\:\:\:{A},{B},{C}\:\:{is}\:{variable} \\ $$$${f}\left({A},{B},{C}\right)=\frac{\sqrt{{cosA}}}{{a}}+\frac{\sqrt{{cosB}}}{{b}}+\frac{\sqrt{{cosC}}}{{c}}= \\ $$$$\frac{\mathrm{1}}{{a}}\left(\sqrt{{cosA}}+{sinA}\left(\frac{\sqrt{{cosB}}}{{sinB}}+\frac{\sqrt{{cosC}}}{{sinC}}\right)\right) \\ $$$${g}\left({A},{B},{C}\right)={A}+{B}+{C}=\mathrm{180} \\ $$$$\Rightarrow\frac{\partial{f}}{\partial{x}_{{i}}…
Question Number 209838 by malwan last updated on 23/Jul/24 $${If}\:{sin}\:{x}\:+\:{cos}\:{x}\:+\:{tan}\:{x}\:+\:{cot}\:{x}\:+ \\ $$$${sec}\:{x}\:+\:{csc}\:{x}\:=\:\mathrm{7}\:,\:{find}\:{sin}\mathrm{2}{x}? \\ $$ Answered by mahdipoor last updated on 23/Jul/24 $$=\frac{{s}^{\mathrm{2}} {c}+{c}^{\mathrm{2}} {s}+{s}^{\mathrm{2}} +{c}^{\mathrm{2}}…
Question Number 209822 by SonGoku last updated on 22/Jul/24 $$\mathrm{A}\:\mathrm{ramp}\:\mathrm{is}\:\mathrm{supported}\:\mathrm{by}\:\mathrm{six}\:\mathrm{pillars}\:\mathrm{and}\:\mathrm{the}\:\mathrm{talleste} \\ $$$$\mathrm{on}\:\mathrm{measures}\:\mathrm{6}\:\mathrm{meters}.\:\mathrm{The}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{eachi} \\ $$$$\mathrm{pllar}\:\mathrm{is}\:\mathrm{5}\:\mathrm{meters}.\:\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{third}\:\mathrm{pillar}?\: \\ $$ Commented by mr W last updated on…
Question Number 209781 by alusto22 last updated on 21/Jul/24 $$\:\:\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°−\mathrm{cos}\:\mathrm{40}°}{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°+\mathrm{cos}\:\mathrm{40}°}\:=? \\ $$ Answered by efronzo1 last updated on 21/Jul/24 $$\:\:\:\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{20}°\right)}{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°+\mathrm{2cos}\:^{\mathrm{2}} \mathrm{20}−\mathrm{1}}\: \\ $$$$\:\:=\:\frac{\mathrm{2sin}\:\mathrm{20}°\left(\mathrm{cos}\:\mathrm{20}°+\mathrm{sin}\:\mathrm{20}°\right)}{\mathrm{2cos}\:\mathrm{20}°\left(\mathrm{cos}\:\mathrm{20}°+\mathrm{sin}\:\mathrm{20}°\right)} \\…
Question Number 209639 by SonGoku last updated on 17/Jul/24 Commented by SonGoku last updated on 17/Jul/24 $$\mathrm{In}\:\mathrm{the}\:\mathrm{figure},\:\mathrm{DC}//\mathrm{FG}\:\mathrm{is}\:\mathrm{the}\:\mathrm{width}\:\mathrm{of}\:\mathrm{a}\:\mathrm{river}. \\ $$$$\mathrm{How}\:\mathrm{wideis}\:\mathrm{this}\:\mathrm{river}? \\ $$$$\left(\mathrm{How}\:\mathrm{do}\:\mathrm{I}\:\mathrm{solve}\:\mathrm{it}?\right) \\ $$ Answered by…
Question Number 209590 by SonGoku last updated on 15/Jul/24 Commented by SonGoku last updated on 15/Jul/24 $$\mathrm{Help}\:-\:\mathrm{me}! \\ $$ Answered by mr W last updated…
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 $$\:\:\frac{\mathrm{1}−\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{6}{sinBcosC}}={tanC} \\…
Question Number 209519 by depressiveshrek last updated on 12/Jul/24 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:{n}\in\mathbb{N}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{sin}\left({n}\right)\in\mathbb{Q}\:? \\ $$ Commented by Rasheed.Sindhi last updated on 12/Jul/24 $${n}\:{in}\:{degrees}? \\ $$ Commented…
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}''…
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