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Category: Trigonometry

Question-210677

Question Number 210677 by Safojon last updated on 16/Aug/24 Answered by BHOOPENDRA last updated on 16/Aug/24 $$\theta=\frac{{k}\pi}{\mathrm{7}\:},\mathrm{7}\theta={k}\pi\:\left({where}\:{k}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}….{n}\right) \\ $$$$\mathrm{7}\theta=\pi\:\left({k}=\mathrm{1}\right) \\ $$$$\mathrm{4}\theta+\mathrm{3}\theta=\pi \\ $$$${tg}\mathrm{4}\theta=−{tg}\left(\mathrm{3}\theta\right)\:{let}\:{tg}={t} \\ $$$$\Rightarrow\:\frac{\mathrm{4}{t}−\mathrm{4}{t}^{\mathrm{3}}…

Question-209880

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}}…

A-ramp-is-supported-by-six-pillars-and-the-talleste-on-measures-6-meters-The-distance-between-eachi-pllar-is-5-meters-What-will-be-the-height-of-the-third-pillar-

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…

1-sin-40-cos-40-1-sin-40-cos-40-

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-209639

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…