Question Number 193635 by mr W last updated on 17/Jun/23 Commented by mr W last updated on 17/Jun/23 $${find}\:{the}\:{largest}\:{circle}\:{and}\:{the}\:{largest} \\ $$$${square}\:{which}\:{you}\:{can}\:{completely} \\ $$$${cover}\:{with}\:{three}\:{circular}\:{plates}\:{with} \\ $$$${radius}\:\mathrm{1}\:{respectively}.…
Question Number 193671 by cherokeesay last updated on 17/Jun/23 Answered by mr W last updated on 17/Jun/23 $${R}={radius} \\ $$$${a}={width}\:{of}\:{rectangle} \\ $$$$\mathrm{4}^{\mathrm{2}} −{a}^{\mathrm{2}} ={a}\left(\mathrm{2}{R}−{a}\right) \\…
Question Number 193522 by Mingma last updated on 16/Jun/23 Answered by Subhi last updated on 16/Jun/23 $$\frac{{x}+{z}}{{sin}\left(\mathrm{100}\right)}=\frac{{x}}{{sin}\left(\mathrm{40}\right)} \\ $$$${sin}\left(\mathrm{100}\right){x}={sin}\left(\mathrm{40}\right){x}+{sin}\left(\mathrm{40}\right){z}\:\Rrightarrow\:{x}=\frac{{sin}\left(\mathrm{40}\right){z}}{{sin}\left(\mathrm{10}\right)−{sin}\left(\mathrm{40}\right)}\:\left({i}\right) \\ $$$$\frac{{z}}{{sin}\left({y}−\mathrm{40}\right)}=\frac{{x}+{z}}{{sin}\left({u}\right)} \\ $$$${u}+{y}−\mathrm{40}=\mathrm{180}−\mathrm{140}=\mathrm{40}\:\Rrightarrow\:{y}=\mathrm{80}−{u}\:\left({ii}\right) \\ $$$$\frac{{z}}{{sin}\left(\mathrm{40}−{u}\right)}=\frac{{x}+{z}}{{sin}\left({u}\right)}…
Question Number 193548 by lmcp1203 last updated on 16/Jun/23 Commented by lmcp1203 last updated on 16/Jun/23 $${hint}\:{by}\:{trigonometry}\:\:{x}=\mathrm{10} \\ $$ Answered by Subhi last updated on…
Question Number 193581 by Rupesh123 last updated on 16/Jun/23 Answered by Subhi last updated on 16/Jun/23 $${a}^{\mathrm{2}} =\left({c}+{d}\right)^{\mathrm{2}} +\left({c}+{e}\right)^{\mathrm{2}} =\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{cd}+\mathrm{2}{ec}+{d}^{\mathrm{2}} +{e}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} ={e}^{\mathrm{2}}…
Question Number 193511 by Lekhraj last updated on 15/Jun/23 Answered by a.lgnaoui last updated on 17/Jun/23 $$\measuredangle\mathrm{OAE}=\alpha\:\:\:\mathrm{OA}=\mathrm{OB}=\mathrm{OF}=\mathrm{R} \\ $$$$\Rightarrow\measuredangle\mathrm{OAF}=\measuredangle\mathrm{OFA}=\alpha \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{AF}}{\mathrm{2R}}=\frac{\mathrm{11}}{\mathrm{2R}}\:\Rightarrow\:\:\:\:\:\boldsymbol{\mathrm{R}}=\frac{\mathrm{11}}{\mathrm{2cos}\:\boldsymbol{\alpha}}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Surface}\:\mathrm{triangle}\:\left(\mathrm{AOE}\right)\:\mathrm{est}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}=\frac{\boldsymbol{\mathrm{OA}}×\boldsymbol{\mathrm{AE}}\mathrm{sin}\:\boldsymbol{\alpha}}{\mathrm{2}}=\frac{\mathrm{11}×\mathrm{4sin}\:\boldsymbol{\alpha}}{\mathrm{cos}\:\boldsymbol{\alpha}}…
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Question Number 193309 by Mingma last updated on 09/Jun/23 Commented by a.lgnaoui last updated on 10/Jun/23 $$\boldsymbol{\mathrm{What}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{question}}\:? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 193293 by TUN last updated on 09/Jun/23 Answered by AST last updated on 09/Jun/23 $${h}={a}+{b}+{c} \\ $$$$\mid{AH}\mid^{\mathrm{2}} =\mid{b}+{c}\mid^{\mathrm{2}} =\frac{\left({b}+{c}\right)^{\mathrm{2}} }{{bc}}=\left({b}+{c}\right)\left(\overset{−} {{b}}+\overset{−} {{c}}\right) \\…
Question Number 193239 by Mingma last updated on 08/Jun/23 Answered by a.lgnaoui last updated on 08/Jun/23 $$\measuredangle\mathrm{B}=\mathrm{180}−\mathrm{110}=\mathrm{70} \\ $$$$\bigtriangleup\mathrm{ABE}\:\:\:\:\measuredangle\mathrm{AEB}=\mathrm{180}−\mathrm{120}=\mathrm{60} \\ $$$$\bigtriangleup\mathrm{ACE}\:\:\:\:\measuredangle\mathrm{AEC}=\mathrm{180}−\mathrm{60}=\mathrm{120} \\ $$$$ \\ $$$$\bigtriangleup\mathrm{ABC}\:\:\:\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{AB}}=\frac{\mathrm{sin}\:\mathrm{70}}{\mathrm{BC}}…