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 168040 by cortano1 last updated on 01/Apr/22

Commented by cortano1 last updated on 01/Apr/22

$$Oiscenterpointofcircle$$

Commented by som(math1967) last updated on 01/Apr/22

$${BC}^2=400{sin}^{2}75$$

Commented by cortano1 last updated on 01/Apr/22

$$igot100(2+\sqrt{3})$$

Commented by som(math1967) last updated on 01/Apr/22

$$sin75=\frac{1}{2}\times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}$$$$=\frac{(\sqrt{3}+1)}{2\sqrt{2}}$$$$400{sin}^{2}75$$$$=400\times \frac{{(\sqrt{3}+1)}^2}{8}$$$$=400\times \frac{2(2+\sqrt{3})}{8}=100(2+\sqrt{3)}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com