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Question-201906




Question Number 201906 by sonukgindia last updated on 15/Dec/23
Answered by AST last updated on 15/Dec/23
sin(x)=(s/h)...(i)  (ii)...((sin(2x))/h)=((sin(x))/(2s))⇒2sin(x)=((sin(x))/(sin(2x)))  ⇒sin(2x)=(1/2)⇒x=15°
$${sin}\left({x}\right)=\frac{{s}}{{h}}…\left({i}\right) \\ $$$$\left({ii}\right)…\frac{{sin}\left(\mathrm{2}{x}\right)}{{h}}=\frac{{sin}\left({x}\right)}{\mathrm{2}{s}}\Rightarrow\mathrm{2}{sin}\left({x}\right)=\frac{{sin}\left({x}\right)}{{sin}\left(\mathrm{2}{x}\right)} \\ $$$$\Rightarrow{sin}\left(\mathrm{2}{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{x}=\mathrm{15}° \\ $$
Answered by mr W last updated on 16/Dec/23
Commented by mr W last updated on 16/Dec/23
AB=BC=2×BE  sin 2α=((BE)/(AB))=(1/2)  ⇒2α=30° ⇒α=15°
$${AB}={BC}=\mathrm{2}×{BE} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha=\frac{{BE}}{{AB}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\alpha=\mathrm{30}°\:\Rightarrow\alpha=\mathrm{15}° \\ $$

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