Category: Geometry

Question Number 185770 by Mingma last updated on 27/Jan/23 Answered by mahdipoor last updated on 27/Jan/23 $$\mathrm{\Delta }ADB:\frac{AB}{sin(180-(20+x))}=\frac{BD}{sin\left(x\right)}$$$$\mathrm{\Delta }BDC:\frac{BD}{sin30}=\frac{BC}{sin(180-(50+30))}$$$$\mathrm{\Delta }ABC:\frac{AB}{sin70}=\frac{BC}{sin(180-(70+70))}$$$$\Rightarrow \Rightarrow BD=\frac{sin\left(x\right)}{sin(20+x)}AB=\frac{sin\left(30\right)}{sin\left(80\right)}BC=$$$$\frac{{sin}\left(\mathrm{30}\right)}{{sin}\left(\mathrm{80}\right)}×\frac{{sin}\left(\mathrm{40}\right)}{{sin}\left(\mathrm{70}\right)}{AB}…

Question Number 54661 by ajfour last updated on 08/Feb/19 Commented by ajfour last updated on 08/Feb/19 $$Given\mathit{a}and\mathit{b},find\mathit{c}.$$ Commented by behi83417@gmail.com last updated on…

Question Number 185685 by cherokeesay last updated on 25/Jan/23 Answered by mr W last updated on 25/Jan/23 $$\frac{r}{2R-r}=\sin 30°=\frac{1}{2}$$$$\Rightarrow \frac{R}{r}=\frac{3}{2}$$$$\frac{{green}}{{yellow}}=\frac{\pi{r}^{\mathrm{2}} }{\pi{R}^{\mathrm{2}} −\pi{r}^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}}…

Question Number 54602 by behi83417@gmail.com last updated on 07/Feb/19 $$inagiventriangle:$$$$\mathbf{tg}\frac{\mathbf{C}}{2}=\frac{\mathbf{a}.\mathbf{tgA}+\mathbf{b}.\mathbf{tgB}}{\mathbf{a}+\mathbf{b}}.$$$$\mathbf{define}\mathbf{the}\mathbf{kind}\mathbf{of}\mathbf{triangle}.$$ Commented by MJS last updated on 08/Feb/19 $$\mathrm{using}\:\mathrm{the}\:\mathrm{laws}\:\mathrm{of}\:\mathrm{sines}\:\mathrm{and}\:\mathrm{cosines}\:\mathrm{I}\:\mathrm{get} \…

Question Number 185637 by Mingma last updated on 24/Jan/23 Answered by HeferH last updated on 24/Jan/23 $$\frac{a}{b}=\frac{1+\sqrt{5}}{2}$$$$\frac{a+b}{\sin (18°+x)}=\frac{a}{\sin 18°}$$$$1+\frac{b}{a}=\frac{\sin (18°+x)}{\sin 18°}$$$$1+\frac{2}{1+\sqrt{5}}=\frac{\sin (18°+x)}{\sin 18°}$$$$\mathrm{sin}^{−\mathrm{1}}…