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 162598 by amin96 last updated on 30/Dec/21

Answered by mr W last updated on 30/Dec/21

Commented by mr W last updated on 30/Dec/21

$$saysidelengthofhexagonis1.$$$$thenthesidelengthofsquareis2.$$$$AK=2-1\times \cos 30°=2-\frac{\sqrt{3}}{2}$$$$\tan \alpha =\frac{KI}{AK}=\frac{1}{2(2-\frac{\sqrt{3}}{2})}=\frac{1}{4-\sqrt{3}}$$$$LE=\frac{AL}{\tan \alpha }=4-\sqrt{3}$$$$ME=LE-2=2-\sqrt{3}$$$$\tan \frac{x}{2}=\frac{BM}{ME}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$$$$\tan x=\frac{2(2+\sqrt{3})}{1-{(2+\sqrt{3})}^2}=-\frac{1}{\sqrt{3}}$$$$\Rightarrow x=180°-{\tan }^{-1}\frac{1}{\sqrt{3}}=180°-30°=150°$$

Commented by amin96 last updated on 30/Dec/21

$$thankssir.greatful$$

Commented by Tawa11 last updated on 30/Dec/21

$$\mathrm{Great}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com