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 50577 by behi83417@gmail.com last updated on 17/Dec/18

Commented by behi83417@gmail.com last updated on 17/Dec/18

$$\measuredangle ABC=90,AC=1,AD=\frac{\sqrt{3}}{6}\left(angularbisector\right).$$$$\Rightarrow AB=?,BC=?$$

Answered by ajfour last updated on 17/Dec/18

$$let\mathrm{\angle }BAC=2\theta ,BC=\sin 2\theta$$$$AB=\cos 2\theta ,AD=\cos 2\theta \sec \theta$$$$AD=\frac{\cos 2\theta }{\cos \theta }=\frac{1}{2\sqrt{3}}$$$$\Rightarrow 2\cos {}^2\theta -1-\frac{\cos \theta }{2\sqrt{3}}=0$$$$4\cos \theta =\frac{1}{2\sqrt{3}}+\sqrt{\frac{1}{12}+8}$$$$=\frac{\sqrt{97}+1}{8\sqrt{3}}$$$$AB=\cos 2\theta =\frac{\cos \theta }{2\sqrt{3}}=\frac{\sqrt{97}+1}{48}$$$$AB\approx 0.226$$$$BC=\sqrt{1-{\left(\frac{\sqrt{97}+1}{48}\right)}^2}\approx 0.974$$

Commented by behi83417@gmail.com last updated on 17/Dec/18

$$thankssirAjfour.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com