Category: Geometry

Question Number 40031 by ajfour last updated on 15/Jul/18 Commented by ajfour last updated on 16/Jul/18 $$CEandDF\mathrm{\perp }AF.Find\theta .$$ Answered by ajfour last updated on…

Question Number 39993 by ajfour last updated on 14/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18 $$let\mathrm{\angle }AOT=\theta$$$$OT\parallel toBCso\frac{AO}{AC}=\frac{AT}{AB}=\frac{OT}{BC}$$$$pointT(cos\theta ,sin\theta )andpointC(-1,0)$$$$eqnofOTisy=xtan\theta$$$${eqn}\:{ofCB}\:{is}\:{y}=\left({x}+\mathrm{1}\right){tan}\theta…

Question Number 105474 by Don08q last updated on 29/Jul/20 Answered by ajfour last updated on 29/Jul/20 Commented by ajfour last updated on 29/Jul/20 $${BD}=\sqrt{\mathrm{25}+\mathrm{36}−\mathrm{2}×\mathrm{5}×\mathrm{6cos}\:\mathrm{30}°} \…

Question Number 39848 by ajfour last updated on 12/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18 $$tan\theta =\frac{a}{1}$$$$sin\theta =\frac{1}{1+a}$$$$tan\theta =\frac{a}{1}sosin\theta =\frac{a}{\sqrt{1+a^2}}$$$$\frac{{a}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}=\frac{\mathrm{1}}{{a}+\mathrm{1}}…

Question Number 39737 by ajfour last updated on 10/Jul/18 Commented by ajfour last updated on 10/Jul/18 $$Sir,$$$$IgotR=\sqrt{1+\frac{1}{\sqrt{2}}}.Pleasejudge,$$$$wrongorright!$$ Commented by…