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Question Number 39737 by ajfour last updated on 10/Jul/18 Commented by ajfour last updated on 10/Jul/18 $${Sir},\: \\ $$$${I}\:{got}\:{R}=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\:\:.{Please}\:{judge}, \\ $$$${wrong}\:{or}\:{right}! \\ $$ Commented by…
Question Number 39678 by ajfour last updated on 09/Jul/18 Commented by ajfour last updated on 09/Jul/18 $$\bigtriangleup{ABC}\:{is}\:{projection}\:{of}\:{an}\: \\ $$$${equilateral}\:\bigtriangleup{APQ}.\:{Find}\:{the}\: \\ $$$${height}\:{p},{q}\:{of}\:{vertices}\:{P}\:{and}\:{Q} \\ $$$${above}\:{ground}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}. \\ $$…
Question Number 39666 by ajfour last updated on 09/Jul/18 Commented by ajfour last updated on 09/Jul/18 $${In}\:{terms}\:{of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{\theta}\:,\:{find}\: \\ $$$$\:\boldsymbol{{x}},\:\boldsymbol{{y}},\:\boldsymbol{\alpha},\:\boldsymbol{\beta},\:{and}\:\boldsymbol{{r}}\:.\:\:\:\left({y}=\:{AD}\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
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Question Number 39626 by ajfour last updated on 08/Jul/18 Commented by ajfour last updated on 09/Jul/18 $${Find}\:\boldsymbol{{b}}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{R}}\:. \\ $$$${What}\:{is}\:\boldsymbol{{a}}\:{in}\:{terms}\:{of}\:\boldsymbol{{R}}\:{if}\:{a}+{b}={R}. \\ $$ Answered by ajfour last…
Question Number 39559 by ajfour last updated on 07/Jul/18 Commented by ajfour last updated on 07/Jul/18 $${Find}\:\boldsymbol{\alpha}\:{and}\:{radius}\:\boldsymbol{{r}}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}}. \\ $$ Answered by MrW3 last updated on…
Question Number 170624 by cortano1 last updated on 27/May/22 Commented by som(math1967) last updated on 28/May/22 Commented by som(math1967) last updated on 28/May/22 Commented by…
Question Number 105057 by I want to learn more last updated on 25/Jul/20 Answered by OlafThorendsen last updated on 25/Jul/20 $$\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\mathrm{2}{a}\right)^{\mathrm{2}} −\mathrm{3}×\frac{\frac{\pi}{\mathrm{3}}}{\mathrm{2}\pi}\pi{a}^{\mathrm{2}} \\ $$$$=\:\left(\sqrt{\mathrm{3}}−\frac{\pi}{\mathrm{2}}\right){a}^{\mathrm{2}} \:\left(\mathrm{4}\right)…
Question Number 39511 by behi83417@gmail.com last updated on 06/Jul/18 Commented by behi83417@gmail.com last updated on 08/Jul/18 $$\boldsymbol{{in}}\:\boldsymbol{{ABCD}},\boldsymbol{{E}},\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{intersect}}\:\boldsymbol{{point}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{diagonals}}.\boldsymbol{{draw}}\:\boldsymbol{{outcircles}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{A}}\overset{\bigtriangleup} {\boldsymbol{{B}E}}\:\boldsymbol{{and}}\:\boldsymbol{{C}}\overset{\bigtriangleup} {\boldsymbol{{D}E}}.\boldsymbol{{say}}\:\boldsymbol{{the}}\:\boldsymbol{{center}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{two}}\:\boldsymbol{{circles}}\:\boldsymbol{{as}}:\boldsymbol{{P}}\:\boldsymbol{{and}}\:\boldsymbol{{Q}}.…