Question

a vertical tower stands on a horizontal plane and is surrounded by a vertical flagstaff of height 'n' .At a point on the plane the angle of elevation of the bottom of the flagstaf is ,'alpha' and that of the top of flagstaff is 'beta' . prove that height of the tower is h tan alpha/tan beta-tan alpha

Answer 1

We have a tower AB of height H .  On top of it, we have a flagstaff BC of height h.  From a point P on the plane
         the angle of elevation of B is APB = α
         the angle of elevation of C is APC = β
    
              tan α = H / AP         tan β = (H + h)  / AP

         =>   h = AP ( tan β - tan α)  = (H / tan α) * (tan β - tan α)

               H = height of tower =  h tan α / ( tan β - tan α )

Answer 2

Question:

A vertical tower stands on a horizontal plane and is surrounded by a vertical flagstaff of height 'n'. At a point on the plane, the angle of elevation of the bottom of the flagstaff is α and that of the top of flagstaff is β.

To Prove:

$\mathrm {Height \ of \ the \ tower \ is \ = \dfrac{htan\alpha}{tan\beta - tan\alpha}}$

Solution:

Height of the Tower $\longrightarrow$ CB

Height of the Flagstaff $\longrightarrow$ DC

Let DC $\longrightarrow$ 'h'

In ΔABD,

∠B = 90°

$\sf tan\beta = \dfrac{Opposite \ Side}{Adjacent \ Side}$

$\sf tan\beta = \dfrac{DB}{AB}$

$\sf tan\beta = \dfrac{h \ + \ CB}{AB}$    → Eq(1)

In ΔABC,

∠B = 90°

$\sf tan \alpha = \dfrac{Opposite \ Side}{Adjacent \ Side}$

$\sf tan\alpha = \dfrac{CB}{AB}$

On cross-multiplying we get,

$\sf AB = \dfrac{CB}{tan\alpha}$      → Eq(2)

We know that,

$\sf tan\beta = \dfrac{h + CB}{AB}$

Substitute the value of AB in the above equation, proved in Eq(2)

$\sf tan\beta = \dfrac{h + CB}{\dfrac{CB}{tan\alpha}}$

$\sf tan\beta = \dfrac{h + CB}{1} \ \times \ \dfrac{tan\alpha}{CB}$

$\sf tan\beta = \dfrac{tan\alpha(h + CB)}{CB}$

On Cross-multiplying we get,

CB × tanβ = tanα(h + CB)

CB × tanβ = htanα + CB × tanα

CBtanβ - CBtanα = htanα

CB(tanβ - tanα) = htanα

$\implies {\large{\boxed{\sf{CB \ = \ \frac{htan\alpha}{tan\beta \ - \ tan\alpha}}}$

Hence Proved.

Note:

Here α and $\alpha$ stand for Alpha.

Here β and $\beta$ stand for Beta.