Question

A person standing at some distance from a tower sees the top of the tower, making an angle of elevation of ${50}^{\circ }$. When he moves 10 m towards the tower the angle of elevation changes to ${70}^{\circ }$. Find the height of the tower.

$[tan{70}^{\circ }=2.74,tan{50}^{\circ }=1.2]$

Answer 1

Let the initial point be B and and BD = 10 m

In Right- angled $△$ ADC

$tan{70}^{\circ }=\frac{AC}{DC}=\frac{AC}{x}$

$AC=x.tan{70}^{\circ }$

$AC=2.74x$ ------------ $\left(i\right)$

In Right- angled $△$ ABC

$tan{50}^{\circ }=\frac{AC}{BC}=\frac{AC}{(x+10)}$

$AC=(x+10).tan{50}^{\circ }$

$AC=(x+10)\times \left(1.2\right)$ -------- $\left(ii\right)$

From i and ii we have

$2.74x=(x+10)1.22.74x=1.2x+121.54x=12x=7.79mAC=xtan{70}^{\circ }=7.79\times 2.74=21.35m\text{Height of tower}=21.35m$