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 124953 by Algoritm last updated on 07/Dec/20

Answered by mr W last updated on 07/Dec/20

say AC=1  ((BC)/(sin 30))=((AC)/(sin 50))  ⇒BC=((sin 30)/(sin 50))  similarly  OC=((sin 10)/(sin 20))  ((sin (α+10))/(sin α))=((BC)/(OC))=((sin 30×sin 20)/(sin 50×sin 10))  cos 10+((sin 10)/(tan α))=((sin 30×sin 20)/(sin 50×sin 10))  tan α=((sin 10)/(((sin 30×sin 20)/(sin 50×sin 10))−cos 10))=(1/( (√3)))  ⇒α=30°

$${say}\:{AC}=\mathrm{1} \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{30}}=\frac{{AC}}{\mathrm{sin}\:\mathrm{50}} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{sin}\:\mathrm{30}}{\mathrm{sin}\:\mathrm{50}} \\ $$$${similarly} \\ $$$${OC}=\frac{\mathrm{sin}\:\mathrm{10}}{\mathrm{sin}\:\mathrm{20}} \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{10}\right)}{\mathrm{sin}\:\alpha}=\frac{{BC}}{{OC}}=\frac{\mathrm{sin}\:\mathrm{30}×\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{50}×\mathrm{sin}\:\mathrm{10}} \\ $$$$\mathrm{cos}\:\mathrm{10}+\frac{\mathrm{sin}\:\mathrm{10}}{\mathrm{tan}\:\alpha}=\frac{\mathrm{sin}\:\mathrm{30}×\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{50}×\mathrm{sin}\:\mathrm{10}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{sin}\:\mathrm{10}}{\frac{\mathrm{sin}\:\mathrm{30}×\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{50}×\mathrm{sin}\:\mathrm{10}}−\mathrm{cos}\:\mathrm{10}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\alpha=\mathrm{30}° \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com