Question Number 41023 by Tawa1 last updated on 31/Jul/18 Commented by maxmathsup by imad last updated on 31/Jul/18 $${drive}\:{before}\:{calculating}… \\ $$ Terms of Service Privacy…
Question Number 172062 by cortano1 last updated on 23/Jun/22 Answered by kapoorshah last updated on 23/Jun/22 $${s}\:=\:\frac{\mathrm{10}\:+\:\mathrm{14}\:+\:\mathrm{16}}{\mathrm{2}} \\ $$$$\:\:\:=\:\mathrm{20} \\ $$$$ \\ $$$${r}\:=\:\sqrt{\frac{\left(\mathrm{20}\:−\:\mathrm{10}\right)\left(\mathrm{20}\:−\:\mathrm{14}\right)\left(\mathrm{20}\:−\:\mathrm{16}\right)}{\mathrm{20}}} \\ $$$$\:\:\:=\:\mathrm{2}\sqrt{\mathrm{3}}…
Question Number 40948 by behi83417@gmail.com last updated on 30/Jul/18 Commented by MJS last updated on 30/Jul/18 $${B}−{C}=\mathrm{2}{A}\:\Rightarrow\:{C}={B}−\mathrm{2}{A} \\ $$$${A}+{B}+{C}=\mathrm{180}°\:\Rightarrow\:{C}=\mathrm{180}°−{A}−{B} \\ $$$${B}−\mathrm{2}{A}=\mathrm{180}°−{A}−{B}\:\Rightarrow\:{A}=\mathrm{2}{B}−\mathrm{180}° \\ $$$$\Rightarrow\:{C}=\mathrm{360}°−\mathrm{3}{B} \\ $$$$…
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Question Number 40867 by ajfour last updated on 28/Jul/18 Commented by MrW3 last updated on 28/Jul/18 $${as}\:{shown}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\frac{{a}}{\mathrm{2}{R}} \\ $$$$\Rightarrow\theta=\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}} \\ $$$${but}\:{I}\:{think}\:{you}\:{meant}\:{something} \\ $$$${others}. \\…
Question Number 106396 by I want to learn more last updated on 05/Aug/20 Commented by john santu last updated on 05/Aug/20 Commented by john santu…
Question Number 40822 by MrW3 last updated on 28/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18 Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18 Commented by…
Question Number 40771 by Tawa1 last updated on 27/Jul/18 Commented by Tawa1 last updated on 27/Jul/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{blue}\:\mathrm{square}\:…\: \\ $$ Answered by MrW3 last updated on…
Question Number 171765 by cherokeesay last updated on 20/Jun/22 Answered by som(math1967) last updated on 21/Jun/22 $${let}\:{OB}={OA}={OM}={r} \\ $$$$\:\therefore{AC}=\sqrt{{r}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} }{\mathrm{4}}}=\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\bigtriangleup{AOC}\sim\bigtriangleup{OLC} \\ $$$$\:\therefore\frac{{LC}}{{OC}}=\frac{\frac{{r}}{\mathrm{2}}}{\frac{{r}\sqrt{\mathrm{5}}}{\mathrm{2}}}…