Question Number 199738 by sonukgindia last updated on 08/Nov/23
Answered by AST last updated on 08/Nov/23
![((sin(x))/(DB))=((sin(α))/(BC));((sin(2x))/(BE=DB))=((sinCEB=sin(90+x))/(BC)) ⇒((DB)/(BC))=((sin(x))/(sin(α)))=((sin(2x))/(sin(90+x))) ⇒sin(α)=((sin(x)[sin90cosx])/(2sinxcosx))=(1/2)⇒α=30°](https://www.tinkutara.com/question/Q199756.png)
$$\frac{{sin}\left({x}\right)}{{DB}}=\frac{{sin}\left(\alpha\right)}{{BC}};\frac{{sin}\left(\mathrm{2}{x}\right)}{{BE}={DB}}=\frac{{sinCEB}={sin}\left(\mathrm{90}+{x}\right)}{{BC}} \\ $$$$\Rightarrow\frac{{DB}}{{BC}}=\frac{{sin}\left({x}\right)}{{sin}\left(\alpha\right)}=\frac{{sin}\left(\mathrm{2}{x}\right)}{{sin}\left(\mathrm{90}+{x}\right)} \\ $$$$\Rightarrow{sin}\left(\alpha\right)=\frac{{sin}\left({x}\right)\left[{sin}\mathrm{90}{cosx}\right]}{\mathrm{2}{sinxcosx}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\alpha=\mathrm{30}° \\ $$
Commented by Mingma last updated on 27/Nov/23
Can you share your solution diagram
Commented by AST last updated on 27/Nov/23
