Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 199738 by sonukgindia last updated on 08/Nov/23

Answered by deleteduser1 last updated on 08/Nov/23

$$\frac{sin\left(x\right)}{DB}=\frac{sin\left(\alpha \right)}{BC};\frac{sin\left(2x\right)}{BE=DB}=\frac{sinCEB=sin(90+x)}{BC}$$$$\Rightarrow \frac{DB}{BC}=\frac{sin\left(x\right)}{sin\left(\alpha \right)}=\frac{sin\left(2x\right)}{sin(90+x)}$$$$\Rightarrow sin\left(\alpha \right)=\frac{sin\left(x\right)\left[sin90cosx\right]}{2sinxcosx}=\frac{1}{2}\Rightarrow \alpha =30°$$

Commented by Mingma last updated on 27/Nov/23

Can you share your solution diagram

Terms of Service

Privacy Policy

Contact: info@tinkutara.com