Question

In the given figure, $AB$ is the diameter of the circle centered at $O.$ If $\angle COA={60}^0,AB=2r,AC=d$ and $CD=l,$ then $l$ is equal to

• A. $d\sqrt{3}$
• B. $\frac{d}{\sqrt{3}}$
• C. $3d$
• D. $\frac{\sqrt{3}d}{2}$

Answer 1

The correct option is A $d\sqrt{3}$

Given:
$AC=d,OA=OB=r,CD=BD=l,\angle COA=\frac{\pi }{3}$
$\therefore A{C}^2=O{A}^2+O{C}^2-2\left(OA\right)\left(OC\right)\cos \frac{\pi }{3}$
$d^2=r^2+r^2-2r\times r\times \cos \frac{\pi }{3}$
$d^2=2r^2-2r^2\times \frac{1}{2}$
$d^2=r^2$
$\therefore d=r$
As angle $\angle COD=\angle BOD$ and $\angle COA={60}^0$
Adding all the angle together will be add upto ${180}^0$
Hence, $\angle COD=\angle BOD={60}^0$
$\Rightarrow \tan {60}^0=\frac{BD}{OB}=\frac{l}{r}\Rightarrow l=r\sqrt{3}=d\sqrt{3}$