Question Number 46637 by rahul 19 last updated on 29/Oct/18 Commented by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18 $${ok}… \\ $$ Commented by rahul 19 last…
Question Number 46636 by azharkhan250963@gmail.com last updated on 29/Oct/18 $$\mathrm{tan}\:\theta=\mathrm{10tan60}^{°} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18 $${tan}\theta=\mathrm{10}×\sqrt{\mathrm{3}}\: \\ $$$${tan}\theta=\mathrm{10}×\mathrm{1}.\mathrm{732} \\ $$$$\theta={tan}^{−\mathrm{1}} \left(\mathrm{17}.\mathrm{32}\right)…
Question Number 177694 by cortano1 last updated on 08/Oct/22 Commented by Strengthenchen last updated on 08/Oct/22 $${as}\:{the}\:{picture}, \\ $$$$\frac{\mathrm{200}}{\mathrm{sin}\:{y}}=\frac{\mathrm{36}}{{sin}\:\mathrm{5}°}\rightarrow\mathrm{sin}\:{y}=\frac{\mathrm{sin}\:\mathrm{5}°×\mathrm{50}}{\mathrm{9}}\rightarrow\mathrm{cos}\:{y}=\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {y}} \\ $$$${x}^{\mathrm{2}} +\mathrm{36}^{\mathrm{2}} −\mathrm{72}{x}\mathrm{cos}\:{y}=\mathrm{200}^{\mathrm{2}} \rightarrow{x}=\frac{\mathrm{72cos}\:{y}\pm\sqrt{\left(\mathrm{72cos}\:{y}\right)^{\mathrm{2}}…
Question Number 112136 by weltr last updated on 06/Sep/20 $${find}\:\:\:\frac{\mathrm{5}}{\mathrm{6}+\mathrm{7sin}\:\mathrm{2}\beta}\:,\:\:\:{if}\:\:\:\mathrm{tan}\:\beta\:\:=\:\:\mathrm{0}.\mathrm{2} \\ $$ Answered by bemath last updated on 06/Sep/20 $$\mathrm{tan}\:\beta\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\rightarrow\begin{cases}{{if}\:\:\beta\:{in}\:{I}−\:{quadrant}\rightarrow\begin{cases}{\mathrm{sin}\:\beta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{26}}}}\\{\mathrm{cos}\:\beta=\frac{\mathrm{5}}{\:\sqrt{\mathrm{26}}}}\end{cases}}\\{{if}\:\beta\:{in}\:{III}−{quadrant}\rightarrow\begin{cases}{\mathrm{sin}\:\beta=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{26}}}}\\{\mathrm{cos}\:\beta=−\frac{\mathrm{5}}{\:\sqrt{\mathrm{26}}}}\end{cases}}\end{cases} \\ $$$$\Leftrightarrow\:\frac{\mathrm{5}}{\mathrm{6}+\mathrm{14sin}\:\beta\mathrm{cos}\:\beta}\:=\:\frac{\mathrm{5}}{\mathrm{6}+\mathrm{14}\left(\frac{\mathrm{5}}{\mathrm{26}}\right)} \\ $$$$=\:\frac{\mathrm{5}×\mathrm{26}}{\mathrm{6}×\mathrm{26}+\mathrm{70}}\:=\:\frac{\mathrm{65}}{\mathrm{78}+\mathrm{35}}\:=\:\frac{\mathrm{65}}{\mathrm{113}} \\…
Question Number 46573 by Rio Michael last updated on 28/Oct/18 $${Show}\:{that}\: \\ $$$${sin}\mathrm{2}{x}\:\equiv\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}} \\ $$ Commented by peter frank last updated on 28/Oct/18 $$\mathrm{sin2x}=\frac{\mathrm{2sinxcosx}}{\mathrm{1}}…
Question Number 46569 by scientist last updated on 28/Oct/18 $${show}\:{that}\:\:{If}\:{a}^{\mathrm{2}} ,{b}^{\mathrm{2}} ,{c}^{\mathrm{2}\:} \:{are}\:{in}\:{A}.{P}\:\:{the}\:{cotA},{cotB},{cotC}\:{are} \\ $$$${also}\:{in}\:{A}.{P} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 28/Oct/18 $$\frac{{sinA}}{{a}}=\frac{{sinB}}{{b}}=\frac{{sinc}}{{c}}={k}\left({say}\right)…
Question Number 46570 by scientist last updated on 28/Oct/18 $${If}\:{in}\:{triangle}\:{ABC}\:\:\:\frac{{cosB}}{{b}}\:=\frac{{cosC}}{{c}},\:{show}\:{that}\:{the} \\ $$$${triangle}\:{is}\:{isosceles} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 28/Oct/18 $$\frac{{sinA}}{{a}}=\frac{{sinB}}{{b}}=\frac{{sinC}}{{c}} \\ $$$$\frac{{cosB}}{{cosC}}=\frac{{sinB}}{{sinC}}=\frac{{b}}{{c}} \\…
Question Number 46568 by scientist last updated on 28/Oct/18 $${show}\:{that}\:{if}\:{the}\:{side}\:{of}\:{a}\:{triangle}\:{are}\:{in}\:{A}.{P}, \\ $$$${then}\:{the}\:{cotangent}\:{also}\:{in}\:{A}.{P} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 112075 by Aina Samuel Temidayo last updated on 06/Sep/20 $$\mathrm{Are}\:\mathrm{there}\:\mathrm{infinitely}\:\mathrm{many}\:\mathrm{solutions}\:\mathrm{to} \\ $$$$\mathrm{sin}\left(\frac{\beta}{\mathrm{2}}\right)=\mathrm{cos}\left(\mathrm{90}−\frac{\beta}{\mathrm{2}}\right)\:? \\ $$ Commented by $@y@m last updated on 06/Sep/20 This is an identity. �� https://en.m.wikipedia.org/wiki/Identity_(mathematics) Commented…
Question Number 112067 by bemath last updated on 06/Sep/20 $$\:\:\sqrt{{bemath}\:−−−−\blacksquare\blacksquare} \\ $$$${find}\:{the}\:{value}\:{of}\:\underset{{m}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\mathrm{tan}\:\left(\frac{{m}\pi}{\mathrm{7}}\right).\: \\ $$ Commented by MJS_new last updated on 06/Sep/20 $$\mathrm{should}\:\mathrm{be}\:=\mathrm{0} \\…