QJEEM17Macbc0: JEE Main 2017 Mathematics - Class 12 JEEM17Macbc0 Solution: Let a vertical tower AB have its end A on the level ground. Let …

Question

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Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If

\angle

BPC =

\beta

, then tan

\beta

is equal to:

Accepted Answer

Let the height of tower

AB = x

and

LCPA = \propto

From the diagram you can see,

\tan \left( { \propto + \beta } \right) = {x \over {2x}} = {1 \over 2}

we know,

\tan \left( { \propto + \beta } \right) = {{\tan \propto + \tan \beta } \over {1 - \tan \propto \tan \beta }}

\therefore\,\,\,

{{\tan \propto + \tan \beta } \over {1 - \tan \propto \tan \beta }} = {1 \over 2}....\left( 1 \right)

From the diagram,

\tan \widehat \propto = {{x/2} \over {2x}} = {1 \over 4}......\left( 2 \right)

Putting value of

\tan \propto

in eq

(1)

,

{{{1 \over 4} + \tan \beta } \over {1 - {1 \over 4}\tan \beta }} = {1 \over 2}

\Rightarrow 1 - {1 \over 4}\tan \beta

= {1 \over 2} + 2\tan \beta

\Rightarrow {{9\tan \beta } \over 4} = {1 \over 2}

\Rightarrow \tan \beta = {2 \over 9}

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