Question Number 68132 by mr W last updated on 05/Sep/19 Commented by kaivan.ahmadi last updated on 05/Sep/19 $${a}+\beta+\theta=\mathrm{180}\:\:\:\left(\mathrm{1}\right) \\ $$$$\alpha+\beta−{a}=\mathrm{0}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{2}\alpha−\mathrm{2}\theta+{b}=\mathrm{0}\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{3}\right)−\left(\mathrm{2}\right):\:\alpha−\mathrm{2}\theta−\beta+\left({a}+{b}\right)=\mathrm{0}\Rightarrow\mathrm{2}\theta+\beta−\alpha=\mathrm{120}\:\:\:\left(\mathrm{4}\right) \\…
Question Number 68110 by TawaTawa last updated on 05/Sep/19 Commented by kaivan.ahmadi last updated on 07/Sep/19 $$\left(\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1}−\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} \Rightarrow{y}=\pm\mathrm{2}\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$${equation}\:{of}\:{BC}: \\ $$$${m}={tg}\mathrm{30}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 68092 by TawaTawa last updated on 04/Sep/19 Commented by TawaTawa last updated on 05/Sep/19 $$\mathrm{I}\:\mathrm{understand}\:\mathrm{very}\:\mathrm{well}\:\mathrm{now}\:\mathrm{sir}. \\ $$ Commented by kaivan.ahmadi last updated on…
Question Number 68041 by A8;15: last updated on 03/Sep/19 Commented by mr W last updated on 03/Sep/19 $${see}\:{Q}#\mathrm{67386} \\ $$ Terms of Service Privacy Policy…
Question Number 2499 by Syaka last updated on 21/Nov/15 $${ABCD}.{EFGH}\:{is}\:{cube},\:{X}\:{is}\:{midpoint}\:{EF} \\ $$$${if}\:{AB}\:=\:\mathrm{6}\:{cm},\:{how}\:{distance}\:{AX}\:{to}\:{BD}?? \\ $$ Commented by prakash jain last updated on 21/Nov/15 $$\mathrm{What}\:\mathrm{is}\:\mathrm{AM}?\:\mathrm{Is}\:\mathrm{X}=\mathrm{M}? \\ $$…
Question Number 2466 by prakash jain last updated on 20/Nov/15 $$\sqrt{−\frac{\mathrm{1}}{\mathrm{5}}\:}−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}=? \\ $$ Answered by Yozzi last updated on 20/Nov/15 $$\sqrt{\frac{\mathrm{1}}{−\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}={i}\sqrt{\frac{\mathrm{1}}{\mathrm{5}}}−\frac{\mathrm{1}}{{i}\sqrt{\mathrm{5}}} \\ $$$$={i}\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{{i}}{\:\sqrt{\mathrm{5}}}=\mathrm{2}\frac{{i}}{\:\sqrt{\mathrm{5}}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}=\frac{\sqrt{\mathrm{1}}}{\:\sqrt{−\mathrm{5}}}=\sqrt{\frac{\mathrm{1}}{−\mathrm{5}}}\Rightarrow\sqrt{\frac{\mathrm{1}}{−\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}\overset{?}…
Question Number 2458 by alib last updated on 20/Nov/15 $${The}\:{medians}\:{of}\:{a}\:{triangle} \\ $$$${are}\:{m}_{\mathrm{1}} ,\:{m}_{\mathrm{2}} ,\:{m}_{\mathrm{3}} . \\ $$$${Find}\:{the}\:{length}\:{of}\:{each}\:{sides}\: \\ $$$${the}\:{triangle}. \\ $$ Answered by prakash jain…
Question Number 67977 by behi83417@gmail.com last updated on 02/Sep/19 Commented by behi83417@gmail.com last updated on 02/Sep/19 $$\boldsymbol{\mathrm{AD}}=\boldsymbol{\mathrm{DC}},\angle\boldsymbol{\mathrm{AEB}}=\mathrm{90}^{\bullet} \:\:. \\ $$$$\boldsymbol{\mathrm{find}}:\:\:\:\boldsymbol{\mathrm{ED}}\:\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}. \\ $$ Answered by $@ty@m123…
Question Number 67973 by behi83417@gmail.com last updated on 02/Sep/19 Commented by behi83417@gmail.com last updated on 02/Sep/19 $$\mathrm{A}\overset{\bigtriangleup} {\mathrm{B}C},\mathrm{is}\:\mathrm{given}. \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}: \\ $$$$\mathrm{1}.\:\:\:\:\:\frac{\boldsymbol{\mathrm{BX}}}{\boldsymbol{\mathrm{XC}}}\:×\:\:\frac{\boldsymbol{\mathrm{CY}}}{\boldsymbol{\mathrm{YA}}}\:×\:\frac{\boldsymbol{\mathrm{AZ}}}{\boldsymbol{\mathrm{ZB}}}=\mathrm{1}. \\ $$$$\mathrm{2}.\:\:\:\:\:\frac{\boldsymbol{\mathrm{PX}}}{\boldsymbol{\mathrm{XA}}}\:+\:\:\frac{\boldsymbol{\mathrm{PY}}}{\boldsymbol{\mathrm{YB}}}\:+\:\frac{\boldsymbol{\mathrm{PZ}}}{\boldsymbol{\mathrm{ZC}}}=\mathrm{1}. \\…
Question Number 67969 by behi83417@gmail.com last updated on 02/Sep/19 $$\boldsymbol{\mathrm{Two}}\:\boldsymbol{\mathrm{triangles}}\:\bigtriangleup_{\mathrm{1}} \:\boldsymbol{\mathrm{and}}\:\bigtriangleup_{\mathrm{2}} \:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}},\boldsymbol{\mathrm{such}}\: \\ $$$$\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}\:\mathrm{1},\boldsymbol{\mathrm{are}}\: \\ $$$$\boldsymbol{\mathrm{equail}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{medians}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}\:\mathrm{2}. \\ $$$$\mathrm{1}.\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ratio}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{areas}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{triangles}}. \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{small}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{of}}\:\bigtriangleup_{\mathrm{1}} ,\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{equail}}\:\boldsymbol{\mathrm{to}}:\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{be}}:\mathrm{90}^{\bullet} . \\…