A vertical pole with height more than `100 m` consists of two parts, the lower being one-third of the whole. At a point on a horizontal plane through the foot and `40 m` from it, the upper part subtends an angle whose tangent is `1/2` What is the height of the pole.

Options

A. `110 m`
B. `200m`
C. `120m`
D. `150m`

Answer

C

Explanation

Let h be the height of pole, upper portion CD subtend angle `theta` at A.
Then, `tan theta = (1)/(2)`
Let lower part BC subtend angle `phi` at A then
In `Delta ABC`,

` tan phi = (BC)/(AB)=(h//3)/(40)=(h)/(120)`
In `Delta ABD`,
`tan (theta + phi)=(BD)/(AB)`
`rArr (tan theta + tan phi)/(1-tan theta tan phi)=(h)/(40)`
`rArr ((1)/(2)+(h)/(120))/(1-(h)/(240))=(h)/(40)`
`rArr (2(60+h))/((240-h))=(h)/(40)`
`rArr 80 (60 + h) = 240 h - h^(2) rArr 4800 + 80 h = 240 h - h^(2)`
`rArr h^(2)-160h + 4800 = 0 rArr (h-120) (h-40)=0`
`rArr h= 120`
[h=40 is discarded, since h `gt` 100 is given]