A vertical pole (more than 50 meters high) consists of two portions. The lower being  13rdof the whole. If the upper portion subtends an angle  tan−112at a point in a horizontal plane through the foot of the pole at a distance 40 meters from it, then the height of the pole is

Let AB = h meter, BC = 2h meter∠APC=α,  ∠APB=β,   ∠BPC=θ so that θ=α−βIt is given that tanθ=12and AP=40m , where we have tanα=3h40,  tanβ=h40Now tanθ=tanα−β=tanα−tanβ1+tanα+tanβ⇒12=3h40−h401+3h40h40=80h1600+3h2⇒3h2−160h+1600=0           ⇒3h−40h−40=0⇒h=40 , and hence 3h=120 is the required length of the pole. Note that  as the pole is given to 3h≠40 be more than 50 meters height.