Question

A vertical tower stands on a horizontal plane is surmounted by a vertical flagstaff of height(h) . At a point on the ground, the angles of elevation of the bottom and the tops of the flagstaff are (alpha) and (beta) respectively. Prove that the height of the tower is
b.tan(alpha) / tan(beta)-tan(alpha)​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us assume that

• Height of tower, AC be 'x' units

and

• Let A be the point on the ground, at a distance of 'y' units away from the base of the tower.

Now, we have

• AC = x units

• CD = h units

• AB = y units

• ∠ ABC = α

• ∠ ABD = β

Consider,

$\rm :\longmapsto\:In \: \triangle \: ABC$

$\rm :\longmapsto\:tan \alpha \: = \dfrac{AC}{AB}$

$\rm :\longmapsto\:tan \alpha \: = \dfrac{x}{y}$

$\bf\implies \:y = \dfrac{x}{tan \alpha } \: - - (1)$

Now,

$\rm :\longmapsto\:In \: \triangle \: ADB$

$\rm :\longmapsto\:tan \beta \: = \dfrac{AD}{AB}$

$\rm :\longmapsto\:tan \beta = \dfrac{h + x}{y}$

$\rm :\longmapsto\:y \: tan \beta \: = x + h$

On Substituting the value of y, we get

$\rm :\longmapsto\:\dfrac{x}{tan \alpha } tan \beta = x + h$

$\rm :\longmapsto\:\dfrac{x}{tan \alpha } tan \beta - x = h$

$\rm :\longmapsto\:x\bigg(\dfrac{tan \beta }{tan \alpha } - 1 \bigg) = h$

$\rm :\longmapsto\:x\bigg( \dfrac{tan \beta \: - \: tan \alpha }{tan \alpha } \bigg) = h$

$\bf\implies \:x = \dfrac{h \: tan \alpha }{tan \beta \: - \: tan \alpha }$

${\boxed{\boxed{\bf{Hence, Proved}}}}$