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`.Find the height of the pole.

Explanation

Let Ab be the and C be the point 2 m apart from B, the base of the pole.

In the above figure `tan theta =1//2.`
In triagle DBC ,
BC= BD cot `alpha`
`therefore 2=H/3cot alpha`
of tan `alpha = H/6`
In trianlge ABC ,
BC= AB cot `beta`
`therefore 2= H cot beta`
or tan `beta =H/2`
Now `tan theta = tan (beta - alpha)`
` therefore 1/2=(tanbeta- tan alpha)/(1+tan beta tan alpha)=(H/2-H/6)/(1+H^2/12)`
`rArr 1+H^2/12=(2H)/3`
`rArr H^(2)-8H+12=0`
`rArr` H=2m or 6 m