Question Number 211651 by MATHEMATICSAM last updated on 15/Sep/24 $$\mathrm{In}\:\mathrm{triangle}\:\mathrm{ABC},\:\angle\mathrm{C}\:=\:\mathrm{60}°.\:\mathrm{If}\:\mathrm{the}\:\mathrm{length} \\ $$$$\mathrm{of}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{of}\:\angle\mathrm{A},\:\angle\mathrm{B}\:\mathrm{and}\:\angle\mathrm{C}\:\mathrm{are} \\ $$$${a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{respectively}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}\:+\:{c}}\:+\:\frac{\mathrm{1}}{{b}\:+\:{c}}\:=\:\frac{\mathrm{3}}{{a}\:+\:{b}\:+\:{c}}\:. \\ $$ Answered by som(math1967) last updated on 15/Sep/24…
Question Number 211609 by MATHEMATICSAM last updated on 14/Sep/24 $$\mathrm{Prove}\:\mathrm{that},\:\mathrm{in}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{ratios}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{sides}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{angles} \\ $$$$\mathrm{are}\:\mathrm{equal}.\:\mathrm{Also}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{each}\:\mathrm{ratio}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circum}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}. \\ $$ Answered by Frix last updated…
Question Number 211575 by BaliramKumar last updated on 13/Sep/24 Answered by som(math1967) last updated on 13/Sep/24 $$\:\left(\boldsymbol{{a}}\right)\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{{m}}+\boldsymbol{{n}}+\frac{\mathrm{1}}{\boldsymbol{{m}}}+\frac{\mathrm{1}}{\boldsymbol{{n}}}\right) \\ $$ Answered by A5T last updated on…
Question Number 211496 by RojaTaniya last updated on 11/Sep/24 Answered by som(math1967) last updated on 11/Sep/24 $$\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{sin}\mathrm{20}{sin}\mathrm{40}+\mathrm{2}{sin}\mathrm{20}{sin}\mathrm{60}+\mathrm{2}{sin}\mathrm{20}{sin}\mathrm{80}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{20}−{cos}\mathrm{60}+{cos}\mathrm{40}−{cos}\mathrm{80}+{cos}\mathrm{60}−{cos}\mathrm{100}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{20}+{cos}\mathrm{40}−{cos}\mathrm{80}+{cos}\mathrm{80}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{cos}\mathrm{30}{cos}\mathrm{10} \\ $$$$={sin}\mathrm{60}{sin}\mathrm{80}…
Question Number 211392 by CrispyXYZ last updated on 08/Sep/24 $$\bigtriangleup{ABC}.\:\mathrm{cos}\:{C}=\frac{\mathrm{sin}\:{A}\:+\:\mathrm{cos}\:{A}}{\mathrm{2}}=\frac{\mathrm{sin}\:{B}\:+\:\mathrm{cos}\:{B}}{\mathrm{2}}. \\ $$$$\mathrm{Find}\:\mathrm{cos}\:{C}. \\ $$ Answered by Frix last updated on 08/Sep/24 $$\mathrm{Rectangular}\:\mathrm{triangle} \\ $$$$\mathrm{cos}\:{A}\:=\mathrm{sin}\:{B}\:\wedge\:\mathrm{sin}\:{A}\:=\:\mathrm{cos}\:{B}\:\Rightarrow\:{C}=\mathrm{90}° \\…
Question Number 211345 by a.lgnaoui last updated on 06/Sep/24 $$\mathrm{E}\boldsymbol{\mathrm{valuer}}:\:\:\frac{\boldsymbol{\mathrm{R}}}{\boldsymbol{\mathrm{r}}\mathrm{1}+\boldsymbol{\mathrm{r}}\mathrm{2}} \\ $$ Commented by a.lgnaoui last updated on 06/Sep/24 Terms of Service Privacy Policy Contact:…
Question Number 211276 by mnjuly1970 last updated on 04/Sep/24 $$ \\ $$$$\:\:{Prove}\:,\:{in}\:{A}\overset{\Delta} {{B}C}\:\::\: \\ $$$$ \\ $$$$\:\:\:\:\:\frac{{cosA}}{{sin}^{\mathrm{2}} {A}}\:+\:\frac{{cosB}}{{sin}^{\mathrm{2}} {B}}\:\:+\frac{{cosC}}{{sin}^{\mathrm{2}} {C}}\:\geqslant\:\frac{{r}}{{R}} \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{r}\::\:{incircle}\:\:{radius} \\…
Question Number 211115 by Durganand last updated on 28/Aug/24 Answered by som(math1967) last updated on 28/Aug/24 $${tanA}+\mathrm{2}{tan}\mathrm{2}{A}+\frac{\mathrm{4}}{{tan}\mathrm{2}.\mathrm{2}{A}} \\ $$$$={tanA}+\mathrm{2}{tan}\mathrm{2}{A}+\frac{\mathrm{4}\left(\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{2}{A}\right)}{\mathrm{2}{tan}\mathrm{2}{A}} \\ $$$$={tanA}+\frac{\mathrm{2}{tan}^{\mathrm{2}} \mathrm{2}{A}+\mathrm{2}−\mathrm{2}{tan}^{\mathrm{2}} \mathrm{2}{A}}{{tan}\mathrm{2}{A}}\: \\…
Question Number 211075 by ajfour last updated on 27/Aug/24 $$\mathrm{sin}\:\left(\theta+\phi\right)=\theta \\ $$$$\mathrm{sin}\:\theta=\phi \\ $$$${find}\:\theta\:{and}\:\phi. \\ $$ Answered by Ghisom last updated on 27/Aug/24 $$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\…
Question Number 210926 by maths_plus last updated on 22/Aug/24 $$\mathrm{valeur}\:\mathrm{de}\::\: \\ $$$$\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\:=\:??? \\ $$ Commented by mr W last updated on 22/Aug/24…