Question

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

• A. $\sqrt{3}l=r=d$
• B. $l=2\sqrt{2}=d\sqrt{2}$
• C. $l=r\sqrt{3}=d\sqrt{3}$
• D. $\sqrt{2}l=r=d$

Answer 1

The correct option is C $l=r\sqrt{3}=d\sqrt{3}$

Given that

$\angle COA={60}^0$

Therefore, $\Rightarrow \angle COB={120}^0$

In $△OBC,\angle OBC={30}^0;OC=r$

Applying sine Rule, we get

$\Rightarrow \frac{r}{\sin {30}^0}=\frac{BC}{\sin {120}^0}$

$\Rightarrow \frac{r}{\frac{1}{2}}=\frac{BC}{\frac{\sqrt{3}}{2}}$

$\Rightarrow BC=\sqrt{3}r$

$\Rightarrow △BDC$ is equilateral triangle as all the angles are ${60}^0$

$\Rightarrow CD=BC=\sqrt{3}r$

$\Rightarrow l=\sqrt{3}r$

$\Rightarrow △AOC$ is equilateral triangle as all angles are ${60}^0$

$\Rightarrow AC=r$

$\Rightarrow d=r$

Therefore, $l=r\sqrt{3}=d\sqrt{3}$