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Category: Geometry

Question-162473

Question Number 162473 by amin96 last updated on 29/Dec/21 Answered by mr W last updated on 29/Dec/21 Commented by amin96 last updated on 29/Dec/21 $$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{alot}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{mr}}\:\boldsymbol{\mathrm{W}}…

2-lines-through-the-point-A-5-1-are-tangent-to-the-circle-x-2-y-2-4x-6y-4-0-Find-the-equation-of-these-2-lines-

Question Number 31249 by Joel578 last updated on 04/Mar/18 $$\mathrm{2}\:\mathrm{lines}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:{A}\left(\mathrm{5},\:\mathrm{1}\right)\:\mathrm{are}\:\mathrm{tangent} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{circle}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\:\mathrm{4}{x}\:+\:\mathrm{6}{y}\:+\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{these}\:\mathrm{2}\:\mathrm{lines} \\ $$ Commented by Joel578 last updated on 04/Mar/18…

Question-162209

Question Number 162209 by Tawa11 last updated on 27/Dec/21 Commented by Rasheed.Sindhi last updated on 27/Dec/21 $$\mathrm{Q}#\mathrm{1} \\ $$$$\mathrm{C}.\:\mathrm{65} \\ $$$$\mathrm{Q}#\mathrm{2} \\ $$$$\mathrm{Which}\:\mathrm{is}\:\alpha\:\:\mathrm{angle}\:\mathrm{in}\:\mathrm{the}\:\mathrm{diagram}? \\ $$…

Question-162071

Question Number 162071 by mr W last updated on 26/Dec/21 Commented by mr W last updated on 27/Dec/21 $${if}\:{the}\:{distances}\:{from}\:{a}\:{point}\:{to}\:{the} \\ $$$${vertices}\:{of}\:{a}\:{triangle}\:{are}\:{p},\:{q},\:{r} \\ $$$${respectively},\:{find}\:{the}\:{maximum}\:{area} \\ $$$${of}\:{the}\:{triangle}\:{and}\:{its}\:{side}\:{lengthes}.…

Question-30862

Question Number 30862 by ajfour last updated on 27/Feb/18 Answered by mrW2 last updated on 27/Feb/18 $$\frac{{a}}{\mathrm{sin}\:{A}}=\frac{{b}}{\mathrm{sin}\:{B}}=\frac{{c}}{\mathrm{sin}\:{C}}=\frac{\mathrm{1}}{\lambda}=\mathrm{2}{R}=\frac{{abc}}{\mathrm{2}\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}} \\ $$$$\frac{{BQ}}{\mathrm{sin}\:\left({C}−\theta\right)}=\frac{{CQ}}{\mathrm{sin}\:\theta}=\frac{{BC}}{\mathrm{sin}\:\left(\pi−\theta−{C}+\theta\right)} \\ $$$$\frac{{BQ}}{\mathrm{sin}\:\left({C}−\theta\right)}=\frac{{CQ}}{\mathrm{sin}\:\theta}=\frac{{a}}{\mathrm{sin}\:{C}} \\ $$$$\Rightarrow{BQ}=\frac{\mathrm{sin}\:\left({C}−\theta\right)}{\mathrm{sin}\:{C}}×{a} \\ $$$$\Rightarrow{CQ}=\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:{C}}×{a}…