Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 70075 by ajfour last updated on 30/Sep/19

Commented by behi83417@gmail.com last updated on 30/Sep/19

Commented by behi83417@gmail.com last updated on 30/Sep/19

$$\measuredangle \mathrm{ACB}=\frac{540}{5}={108}^{\bullet }$$$${\mathrm{AB}}^2={\mathrm{a}}^2+{\mathrm{a}}^2-{2\mathrm{a}}^2{\cos 108}^{\bullet }$$$${\mathrm{AB}}^2={\mathrm{BH}}^2+1^2$$$$\Rightarrow {2\mathrm{a}}^2-{2\mathrm{a}}^2\cos 108-\frac{{\mathrm{a}}^2}{4}=1$$$$\Rightarrow {7\mathrm{a}}^2-{8\mathrm{a}}^2\cos 108=4$$$$\Rightarrow {\mathrm{a}}^2=\frac{4}{7-8\mathbf{cos}{108}^{\bullet }}\Rightarrow \mathrm{a}=\frac{2}{\sqrt{7-{8\cos 108}^{\bullet }}}$$$${\cos 108}^{\bullet }=\frac{1-\sqrt{5}}{4}$$$$\Rightarrow \mathbf{a}=\frac{2}{\sqrt{7-8\frac{1-\sqrt{5}}{4}}}=\frac{2}{\sqrt{5+2\sqrt{5}}}.[=0.65]$$

Commented by ajfour last updated on 01/Oct/19

$$allanglesmaynotbeequal,i$$$$think..$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com