Consider a semicircle of radius 1 unit constructed on the diameter $\mathrm{AB}$, and let $\mathrm{O}$ be its centre. Let $\mathrm{C}$ be a point on $\mathrm{AO}$ such that $\mathrm{AC}: \mathrm{CO}=2: 1$. Draw $\mathrm{CD}$ perpendicular to $\mathrm{AO}$ with $\mathrm{D}$ on the semicircle. Draw $\mathrm{OE}$ perpendicular to $\mathrm{AD}$ with $\mathrm{E}$ on AD. Let $\mathrm{OE}$ and $\mathrm{CD}$ intersects at $\mathrm{H}$. Then DH equals

• A

  $\frac{1}{\sqrt{5}}$
• B

  $\frac{1}{\sqrt{3}}$
• Correct answer
  C

  $\frac{1}{\sqrt{2}}$
• D

  $\frac{\sqrt{5}-1}{2}$

Solution

'E' is mid point of $A D$. $\begin{array}{l} \cos 2 \theta=\frac{1}{3} \quad \Rightarrow 2 \cos ^{2} \theta-1=\frac{1}{3} \\ \Rightarrow \cos ^{2} \theta=\frac{2}{3} \\ \Rightarrow \cos \theta=\frac{\sqrt{2}}{\sqrt{3}} \\ \sin \theta=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}} \end{array}$ $\frac{\text { ED }}{1}=\frac{1}{\sqrt{3}}$ $\mathrm{DH}=\mathrm{ED} \sec \theta$ $=\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{2}}$ $=\frac{1}{\sqrt{2}}$

Hence, the correct answer is option: 'C'.