Question Number 193423 by mustafazaheen last updated on 13/Jun/23 $$\mathrm{when}\:\:\:\mathrm{tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{a}} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{cos}\theta=?\:\mathrm{from}\:\mathrm{the}\:\mathrm{a} \\ $$ Answered by AST last updated on 13/Jun/23 $${tan}\left(\frac{\theta}{\mathrm{2}}+\frac{\theta}{\mathrm{2}}\right)=\frac{\mathrm{2}{tan}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}\Rightarrow{tan}\left(\theta\right)=\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} −\mathrm{1}} \\…
Question Number 193399 by Mingma last updated on 12/Jun/23 Answered by MathematicalUser2357 last updated on 10/Sep/23 $${A}=\mathrm{cosec}\:\mathrm{39}°\approx\mathrm{1}.\mathrm{589016} \\ $$$${B}=\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{618034} \\ $$$$\mathrm{So},\:{A}\rangle{B} \\ $$ Terms of…
Question Number 193390 by byaw last updated on 12/Jun/23 $$ \\ $$A ship X sailing with a velocity (21 kmh 052⁰) observes a light from…
Question Number 193385 by AnshKumar last updated on 12/Jun/23 Answered by som(math1967) last updated on 12/Jun/23 $${L}.{H}.{S} \\ $$$$=\frac{{sec}\mathrm{8}{A}−\mathrm{1}}{{sec}\mathrm{4}{A}−\mathrm{1}} \\ $$$$=\frac{\frac{\mathrm{1}}{{cos}\mathrm{8}{A}}−\mathrm{1}}{\frac{\mathrm{1}}{{cos}\mathrm{4}{A}}−\mathrm{1}} \\ $$$$=\frac{{cos}\mathrm{4}{A}\left(\mathrm{1}−{cos}\mathrm{8}{A}\right)}{{cos}\mathrm{8}{A}\left(\mathrm{1}−{cos}\mathrm{4}{A}\right)} \\ $$$$=\frac{{cos}\mathrm{4}{A}×\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 193381 by Mingma last updated on 12/Jun/23 Answered by som(math1967) last updated on 12/Jun/23 $$\:\mathrm{2}{sin}^{\mathrm{2}} \mathrm{4}\theta\:+\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{2}{sin}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\mathrm{1}−{cos}\mathrm{8}\theta+\mathrm{1}−{cos}\mathrm{4}\theta−\mathrm{1}+{cos}\mathrm{2}\theta=\mathrm{1} \\ $$$${cos}\mathrm{2}\theta−\left({cos}\mathrm{4}\theta+{cos}\mathrm{8}\theta\right)=\mathrm{0} \\…
Question Number 193366 by aipman last updated on 11/Jun/23 $$\mathrm{2sin}\:^{\mathrm{2}} \mathrm{2}{x}>\mathrm{3cos}\:{x}+\mathrm{3} \\ $$ Answered by Frix last updated on 11/Jun/23 $$\mathrm{2sin}^{\mathrm{2}} \:\mathrm{2}{x}\:−\mathrm{3}\left(\mathrm{1}+\mathrm{cos}\:{x}\right)>\mathrm{0} \\ $$$${c}=\mathrm{cos}\:{x} \\…
Question Number 193349 by Mingma last updated on 11/Jun/23 Answered by som(math1967) last updated on 11/Jun/23 $$\angle{P}=\angle{S} \\ $$$$\angle{Q}=\mathrm{18}\:\therefore\angle{P}=\mathrm{90}−\mathrm{18}=\mathrm{72} \\ $$$$\therefore\angle{S}=\mathrm{72} \\ $$ Commented by…
Question Number 193266 by Mingma last updated on 09/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 193205 by Rupesh123 last updated on 07/Jun/23 Commented by a.lgnaoui last updated on 08/Jun/23 Answered by a.lgnaoui last updated on 07/Jun/23 $$\bigtriangleup\boldsymbol{\mathrm{ABC}}\:\:\:\:\measuredangle\boldsymbol{\mathrm{ABC}}=\boldsymbol{\alpha}\:\:\:\measuredangle\boldsymbol{\mathrm{ACB}}=\boldsymbol{\beta} \\…
Question Number 193203 by Mingma last updated on 07/Jun/23 Answered by som(math1967) last updated on 07/Jun/23 $$\frac{{DB}}{{DC}}=\frac{{BE}}{{EC}}\:\:\left[{DE}\:{is}\:{bisector}\:{of}\angle{BDC}\right] \\ $$$$\:\frac{{DB}}{{DC}}=\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$${Sin}\angle{DCB}={Sin}\mathrm{30} \\ $$$$\angle{DCB}=\mathrm{30} \\ $$$$\:\frac{{DB}}{{BC}}={tan}\mathrm{30}…