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

Question

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

Infinity Learn · Accepted Answer

Let AB $$=$$ h meter, BC $$=$$ $$2h$$ meter∠$$APC=$$α,  ∠$$APB=$$β,   ∠$$BPC=$$θ so that θ$$=$$α−βIt is given that $$tan$$θ$$=12and$$ $$AP=40m$$ , where we have $$tan$$α$$=3h40$$,  $$tan$$β$$=h40Now$$ $$tan$$θ$$=$$$$tan$$α−β$$=$$$$tan$$α−$$tan$$β$$1+$$$$tan$$α$$+$$$$tan$$β⇒$$12=3h40$$−$$h401+3h40h40=80h1600+3h2$$⇒$$3h2$$−$$160h+1600=0$$           ⇒$$3h$$−$$40h$$−$$40=0$$⇒$$h=40$$ , 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.

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

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