Question Number 113849 by ShakaLaka last updated on 15/Sep/20 $$\mathrm{A}\:\mathrm{cricket}\:\mathrm{club}\:\mathrm{has}\:\mathrm{15}\:\mathrm{members},\:\mathrm{of}\:\mathrm{whom} \\ $$$$\mathrm{only}\:\mathrm{5}\:\mathrm{can}\:\mathrm{bowl}.\:\mathrm{If}\:\:\mathrm{the}\:\mathrm{names}\:\mathrm{of}\:\mathrm{15} \\ $$$$\mathrm{members}\:\mathrm{are}\:\mathrm{put}\:\mathrm{into}\:\mathrm{a}\:\mathrm{box}\:\mathrm{and}\:\mathrm{11}\:\mathrm{are} \\ $$$$\mathrm{drawn}\:\mathrm{at}\:\mathrm{random},\:\mathrm{then}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{of}\:\mathrm{obtaining}\:\mathrm{an}\:\mathrm{11}\:\mathrm{containing}\:\mathrm{at}\:\mathrm{least} \\ $$$$\mathrm{3}\:\mathrm{bowlers}\:\mathrm{is} \\ $$ Terms of Service…
Question Number 113846 by deepraj123 last updated on 15/Sep/20 $$\mathrm{Thr}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{tan}\:\mathrm{3}{x}=\mathrm{tan}\:\mathrm{5}{x}\:\:\mathrm{is} \\ $$ Answered by MJS_new last updated on 15/Sep/20 $${x}=\mathrm{0}+{n}\pi={n}\pi;\:{n}\in\mathbb{Z} \\ $$ Answered…
Question Number 113847 by deepraj123 last updated on 15/Sep/20 $$\mathrm{The}\:\mathrm{two}\:\mathrm{adjacent}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{cyclic} \\ $$$$\mathrm{quadrilateral}\:\mathrm{are}\:\mathrm{2}\:\mathrm{and}\:\mathrm{5}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{angle}\:\mathrm{between}\:\mathrm{them}\:\mathrm{is}\:\mathrm{60}°.\:\mathrm{If}\:\mathrm{the}\:\mathrm{third} \\ $$$$\mathrm{side}\:\mathrm{is}\:\mathrm{3},\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{fourth}\:\mathrm{side}\:\mathrm{is} \\ $$ Answered by 1549442205PVT last updated on 16/Sep/20…
Question Number 113842 by deepraj123 last updated on 15/Sep/20 $$\mathrm{If}\:\:\mathrm{in}\:\mathrm{a}\:\mathrm{triangle}\: \\ $$$$\:\:\:{a}\:\mathrm{cos}^{\mathrm{2}} \left(\frac{{C}}{\mathrm{2}}\right)+{c}\:\mathrm{cos}^{\mathrm{2}} \left(\frac{{A}}{\mathrm{2}}\right)=\frac{\mathrm{3}{b}}{\mathrm{2}}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{are}\:\mathrm{in} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 113841 by deepraj123 last updated on 15/Sep/20 $$\mathrm{In}\:\mathrm{any}\:\bigtriangleup{ABC},\:\:{b}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}{C}+{c}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}{B}\:= \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 113824 by deepraj123 last updated on 15/Sep/20 $$\mathrm{If}\:\mathrm{4}\:\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{cos}^{−\mathrm{1}} {x}=\:\pi,\:\mathrm{then}\:{x}\:\mathrm{equals} \\ $$ Answered by $@y@m last updated on 15/Sep/20 $${Let}\:\mathrm{sin}\:^{−\mathrm{1}} {x}=\alpha\:\Rightarrow\mathrm{sin}\:\alpha={x}\:…\left(\mathrm{1}\right) \\ $$$${Let}\:\mathrm{cos}\:\:^{−\mathrm{1}}…
Question Number 113825 by deepraj123 last updated on 15/Sep/20 $$\mathrm{The}\:\mathrm{greatest}\:\mathrm{and}\:\mathrm{least}\:\mathrm{values}\:\mathrm{of} \\ $$$$\left(\mathrm{sin}^{−\mathrm{1}} {x}\right)^{\mathrm{3}} +\:\left(\mathrm{cos}^{−\mathrm{1}} {x}\right)^{\mathrm{3}} \:\:\mathrm{are} \\ $$ Answered by MJS_new last updated on 15/Sep/20…
Question Number 113822 by deepraj123 last updated on 15/Sep/20 $$\mathrm{If}\:\:\mathrm{in}\:\mathrm{a}\:\bigtriangleup{ABC},\:\mathrm{3}{a}={b}+{c},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:\mathrm{is} \\ $$ Answered by $@y@m last updated on 15/Sep/20 $$\mathrm{cot}\:^{\mathrm{2}} \frac{{B}}{\mathrm{2}}=\frac{{s}\left({s}−{b}\right)}{\left({s}−{a}\right)\left({s}−{c}\right)}\:\: {Formula} \\…
Question Number 113823 by deepraj123 last updated on 15/Sep/20 $$\mathrm{A}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+{x}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)\:=\:\frac{\pi}{\mathrm{2}}\:\:\mathrm{is} \\ $$ Answered by MJS_new last updated on 15/Sep/20 $${x}=\mathrm{0}\:\mathrm{because}\:\mathrm{arctan}\:\mathrm{1}\:=\frac{\pi}{\mathrm{4}} \\…
Question Number 113808 by deepraj123 last updated on 15/Sep/20 $$\mathrm{In}\:\mathrm{a}\:\bigtriangleup{ABC}\:,\:\angle{B}=\frac{\pi}{\mathrm{3}}\:\mathrm{and}\:\angle{C}=\frac{\pi}{\mathrm{4}}.\:\mathrm{Let} \\ $$$${D}\:\mathrm{divide}\:{BC}\:\mathrm{internally}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio} \\ $$$$\mathrm{1}\::\:\mathrm{3}.\:\mathrm{Then}\:\frac{\mathrm{sin}\:\angle{BAD}}{\mathrm{sin}\:\angle{CAD}}\:\mathrm{equals} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com