Question Number 44480 by peter frank last updated on 29/Sep/18 $${prove}\:{that}\:\:\frac{\mathrm{9}\pi}{\mathrm{8}\:\:}−\frac{\mathrm{9}}{\mathrm{4}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{9}}{\mathrm{4}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$ Answered by math1967 last updated on 30/Sep/18 $${L}.{H}.{S}=\frac{\mathrm{9}}{\mathrm{4}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\right) \\…
Question Number 44436 by ajfour last updated on 29/Sep/18 Commented by ajfour last updated on 29/Sep/18 $${Find}\:{R}\:{in}\:{terms}\:{of}\:{r}. \\ $$ Answered by MrW3 last updated on…
Question Number 175237 by nadovic last updated on 24/Aug/22 $$\mathrm{The}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:\:\mathrm{AC},\:\mathrm{BC} \\ $$$$\mathrm{and}\:\mathrm{AB}\:\mathrm{of}\:\:\mathrm{a}\:\mathrm{right}−\mathrm{angled}\:\mathrm{triangled} \\ $$$$\mathrm{with}\:\mathrm{lengths}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:{y}\:=\:−\mathrm{7}, \\ $$$${x}=\mathrm{11}\:\mathrm{and}\:\mathrm{4}{x}−\mathrm{3}{y}−\mathrm{5}=\mathrm{0}\:\mathrm{respectively}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{inscribed} \\ $$$$\mathrm{circle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle},\:\mathrm{if}\:\mathrm{its}\:\mathrm{radius}\:{r}, \\ $$$$\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:{r}\:=\:\frac{{a}+{b}−{c}}{\mathrm{2}}. \\ $$ Terms…
Question Number 44117 by peter frank last updated on 21/Sep/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 44096 by peter frank last updated on 21/Sep/18 Answered by $@ty@m last updated on 21/Sep/18 $$\left({a}\right)\:\mathrm{cos}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}}…
Question Number 44088 by peter frank last updated on 21/Sep/18 Answered by $@ty@m last updated on 21/Sep/18 $$\left(\boldsymbol{{a}}\right)\:\angle{BCA}=\angle{BCD}\:\left({same}\:{angle}\right) \\ $$$$\angle{ABC}=\angle{BDC}\:\left({right}\:{angle}\right) \\ $$$$\angle{BAC}=\angle{DBC}\:\left({third}\:{angle}\right) \\ $$$$\therefore\bigtriangleup{ABC}\sim\bigtriangleup{BDC} \\…
Question Number 43990 by ajfour last updated on 19/Sep/18 Commented by ajfour last updated on 27/Sep/18 $${Unique}\:{parabola}\:{or}\:{not}\:? \\ $$$${Asking}\:{again}.\:{some}\:{batsmen} \\ $$$${say},\:{not}\:{unique},\:{i}\:{seek}\:{opinions} \\ $$$${again}.. \\ $$…
Question Number 109483 by ajfour last updated on 24/Aug/20 Commented by ajfour last updated on 24/Aug/20 $${If}\:{the}\:{cubic}\:{curve}\:{has}\:{equation}, \\ $$$${y}={x}^{\mathrm{3}} −\mathrm{21}{x}−\mathrm{20} \\ $$$${Find}\:{coordinates}\:{of}\:{P}. \\ $$ Commented…
Question Number 43921 by peter frank last updated on 17/Sep/18 $${prove}\:{that}\:{product}\:{of}\:{lengths}\:{of}\:{perpendiculars} \\ $$$${from}\:{any}\:{point}\:{of}\:{hyperbola}\:{to}\:{its} \\ $$$${asymptotes}\:{is}\:{constant} \\ $$ Answered by math1967 last updated on 18/Sep/18 $${let}\:{equn}.\:{of}\:{hyperbola}\:{is}\frac{{x}^{\mathrm{2}}…
Question Number 43919 by peter frank last updated on 17/Sep/18 $${The}\:{tangent}\:{at}\:{P}\:{to}\:{an}\:{ellipse}\:{meet}\: \\ $$$${diretrix}\:{at}\:{Q}.{prove}\:{that}\:{the}\:{line}\:{joining} \\ $$$${the}\:{corresponding}\:{focus}\:{to}\:{P}\:{and}\:{Q}\:{are}\:{perpendicular} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com