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 BPC = , then tan is equal to:

Let the height of tower and From the diagram you can see, we know, From the diagram, Putting value of in eq ,

Let a vertical tower AB have its end A on the level ground.... | JEE Main 2020 Height and Distance

Q: JEE Main 2020 Height and Distance: 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 ...

The correct answer is A. Let the height of tower and From the diagram you can see, we know, From the diagram, Putting value of in eq ,

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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