Question Number 26772 by Rasheed.Sindhi last updated on 29/Dec/17 $$\mathrm{With}\:\mathrm{a}\:\mathrm{center}\:\boldsymbol{\mathrm{on}}\:\mathrm{a}\:\mathrm{given}\:\mathrm{circle}\:\mathrm{of} \\ $$$$\mathrm{radius}\:\mathrm{r}\:,\mathrm{an}\:\mathrm{arc}\:\mathrm{has}\:\mathrm{been}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{order} \\ $$$$\mathrm{to}\:\mathrm{divide}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{two}\:\mathrm{equal} \\ $$$$\left(\mathrm{in}\:\mathrm{area}\right)\:\mathrm{parts}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:\:\mathrm{r}\:\left(\mathrm{radius}\:\mathrm{of}\:\mathrm{given}\:\mathrm{circle}\right)?\: \\ $$ Answered by mrW1…
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Question Number 92269 by Power last updated on 05/May/20 Answered by mr W last updated on 05/May/20 $${let}\:{BD}=\mathrm{1} \\ $$$$\frac{{AB}}{\mathrm{sin}\:\mathrm{51}}=\frac{{BD}}{\mathrm{sin}\:\left(\mathrm{25}+\mathrm{51}\right)} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{sin}\:\mathrm{51}}{\mathrm{sin}\:\mathrm{76}} \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{99}}=\frac{{BD}}{\mathrm{sin}\:\left(\mathrm{15}+\mathrm{99}\right)} \\…
Question Number 26723 by Rasheed.Sindhi last updated on 28/Dec/17 $$\mathcal{T}{ake}\:{a}\:{point}\:\boldsymbol{{on}}\:{a}\:{given}\:{circle} \\ $$$${as}\:{a}\:{center}\:{and}\:{draw}\:{an}\:{arc} \\ $$$${which}\:{divide}\:{the}\:{given}\:{circle} \\ $$$${into}\:{two}\:{equal}\left(\mathrm{in}\:\mathrm{area}\right)\:{regions}.{Use}\:{only} \\ $$$${Eucledean}\:{tools}. \\ $$ Commented by prakash jain last…
Question Number 26711 by ktomboy1992 last updated on 28/Dec/17 $$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{right}\:\mathrm{tringle}.\mathrm{prove}\:\mathrm{it}? \\ $$ Commented by prakash jain last updated on 28/Dec/17 $$\mathrm{did}\:\mathrm{u}\:\mathrm{forget}\:\mathrm{to}\:\mathrm{add}\:\mathrm{a}\:\mathrm{picture}\:\mathrm{as}\:\mathrm{comment}? \\ $$ Terms of…
Question Number 157739 by gsk2684 last updated on 27/Oct/21 $${two}\:{parallel}\:{paths}\:\mathrm{26}{m}\:{apart}\:{run} \\ $$$${east}−{west}\:{through}\:{the}\:{woods}. \\ $$$${one}\:{person}\:{east}\:{on}\:{one}\:{path}\:{at}\: \\ $$$$\mathrm{3}\:{kmph}\:{and}\:{another}\:{west}\:{on}\:{other}\: \\ $$$${path}\:{at}\:\mathrm{4}\:{kmph}.\:{if}\:{they}\:{pass}\:{each}\: \\ $$$${other}\:{at}\:{time}\:{t}\:=\mathrm{0}\:,\: \\ $$$$\left.{i}\right){how}\:{far}\:{apart}\:{are}\:{they} \\ $$$$\mathrm{8}\:{sec}\:{later}?\: \\…
Question Number 26652 by ajfour last updated on 27/Dec/17 Commented by ajfour last updated on 27/Dec/17 $${If}\:\:{length}\:\boldsymbol{{c}}\:{is}\:{given}\:{and}\:{it}\:{is} \\ $$$${desired}\:{to}\:{start}\:{from}\:{A}\:{and} \\ $$$${reach}\:{B}\:{along}\:{APB}\:{then}\:{at} \\ $$$${what}\:{angle}\:\boldsymbol{\theta}\:{must}\:{we}\:{proceed}; \\ $$$$\left({construction}\:{method}\:{i}\:{mean}\right).…
Question Number 26644 by tawa tawa last updated on 27/Dec/17 Answered by Rasheed.Sindhi last updated on 27/Dec/17 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{square}=\mathrm{x} \\ $$$$\mathrm{The}\:\mathrm{diagonal}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }=\mathrm{28}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{2}}\:\:\mathrm{x}=\mathrm{28}\sqrt{\mathrm{2}} \\…
Question Number 157652 by ajfour last updated on 26/Oct/21 Commented by ajfour last updated on 26/Oct/21 $${just}\:{tried}\:{a}\:{symmetrical}\:{eq}.\:{plot}! \\ $$ Commented by mr W last updated…
Question Number 26389 by ajfour last updated on 24/Dec/17 Commented by ajfour last updated on 24/Dec/17 $$\:\:\:{r}+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }={c}\:;\:\:{or}\:\:\:\mathrm{cos}\:\theta+\mathrm{sec}\:^{\mathrm{2}} \theta\:={c} \\ $$$$\Rightarrow\:\:{if}\:\:{x}=\frac{\mathrm{1}}{{r}}\:\:\:{then}\:\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}}={c}\: \\ $$$${or}\:\:\:{x}^{\mathrm{3}} −{cx}+\mathrm{1}=\mathrm{0}…