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 177205 by HeferH last updated on 02/Oct/22

Answered by mr W last updated on 02/Oct/22

$$\mathrm{\angle }CAD=180-120-3x=60-3x$$$$\mathrm{\angle }CDB=180-3x$$$$applyinglawofsines:$$$$\frac{\sin (60-3x)}{CD}=\frac{\sin 120}{AD}...\left(i\right)$$$$\frac{CD}{\sin 2x}=\frac{CB}{\sin (180-3x)}...\left(ii\right)$$$$AD=CB$$$$\left(i\right)\times \left(ii\right):$$$$\frac{\sin (60-3x)}{\sin 2x}=\frac{\sin 120}{\sin (180-3x)}$$$$\frac{\sin (60-3x)}{\sin 2x}=\frac{\sqrt{3}}{2\sin 3x}$$$$\Rightarrow x\approx 8.02°$$

Commented by Tawa11 last updated on 02/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com