A man standing on a level plane observes the elevation of the top of a pole to be `theta`. He then walks a distance equal to double the height of the pole and then finds that the elevation is now `2theta`. The value of `cottheta` is

Options

A. `sqrt(2)+1`
B. `2-sqrt(3)/(2)`
C. `sqrt(2-1)`
D. `2+sqrt(3)`

Explanation

Let Ab be the pole of height h, Also let D and C be the initial and final positions of the man .

In triangle ABN,
`x= h cot 2 theta`
In triangle ABD
`h=(2 h + x) tan theta `
Put the value of x from (1) into (2) we get
`1=(2+1/(tan 2 theta)`
`rArr 1=(2+(1-tan ^2 theta )/(2 tan theta )) tan theta`
`therefore tan ^2 theta-4tantheta+1=0` ,
`rArr tantheta = 2+-sqrt3`
`tantheta =2+sqrt3 ("Rejected as other wise" 2theta=150^@) `
`therefore tantheta = (2-sqrt3) `
`rArr theta = 15^@`