Let a vertical tower AB have its end A on the level ground.

Question

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$(1)$,

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

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

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

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

Solutions · 1