question_answer

ABCD is a square. The diagonals A C and BD meet at O. Let K, L be the points on AB such that AO = AK, BO = BL. If \[\theta =\angle LOK,\] then what is the value of \[\tan \theta ?\]

A)  \[\frac{1}{\sqrt{3}}\]                            

B)  \[\sqrt{3}\]

C)  1        

D)  \[\frac{1}{2}\]

Correct Answer: C

Solution :

Let sides of a square be a.

Then,    \[AC=a\sqrt{2}\] and \[AO=OC=\frac{a}{\sqrt{2}}\]

Here,     \[AM=\frac{a}{2}\]

\[\therefore \]      \[LM=\frac{a}{\sqrt{2}}-\frac{a}{2}\] and \[OM=\frac{a}{2}\]

In \[\Delta OML,\]

\[\tan \frac{\theta }{2}=\frac{\frac{a}{\sqrt{2}}-\frac{a}{2}}{\frac{a}{2}}=\frac{\frac{\sqrt{2}-1}{2}}{\frac{1}{2}}=\sqrt{2}-1\]

\[\therefore \]      \[\tan \theta =\frac{2\tan \frac{\theta }{2}}{1-{{\tan }^{2}}\frac{\theta }{2}}=\frac{2(\sqrt{2}-1)}{1-(2+1-2\sqrt{2})}\]

\[=\frac{2(\sqrt{2}-1)}{1-3+2\sqrt{2}}=\frac{2\,\,(\sqrt{2}-1)}{2\sqrt{2}-2}\]\[\Rightarrow \]\[\tan \theta =1\]