Question

From top of a tower the angle of depression of top and bottom of a multistored building are ${30}^o$ and ${60}^o$ respectively. If the height of the building is 100m. find the height if a tower.

Answer 1

Let $AB=$height of the building$=100m$

$DE=AC=AB+BC=h=$ height of the tower

$\angle ADC={60}^{\circ }$

$\angle BDC={60}^{\circ }$

Now

$\tan {30}^{\circ }=\frac{CB}{CD}=\frac{x}{d}$

$\frac{1}{\sqrt{3}}=\frac{x}{d}$ (1)

$\tan {60}^{\circ }=\frac{CA}{CD}=\frac{x+100}{d}$

$\sqrt{3}=\frac{x+100}{d}$ (2)

From (1)

$d=\sqrt{3}x$ (3)

Substituting (3) in (2)

$\sqrt{3}=\frac{(x+100)}{\sqrt{3}x}$

$3x=x+100$

$2x=100$

$x=50$

Height of the tower$=DE=h=AB+CB=100m+x=100m+50m=150m$