Question

From the top of a building $50\sqrt{3}m$ high, the angle of depression of an object on the ground is observed to be ${45}^o$. Find the distance of the object from the building.

Answer 1

Let $△ABC$ be a right angled triangle where $\angle B={90}^0$ and $\angle DAC={45}^0$ as shown in the above figure.

Let $x$ be the distance of object from the building.

Since $AD||BC$, therefore,

$\angle DAC=\angle BCA={45}^0$ (Alternate angles)

We know that $\tan \theta =\frac{Oppositeside}{Adjacentside}=\frac{AB}{BC}$

Here, $\theta ={45}^0$, $BC=x$ m and $AB=50\sqrt{3}$ m, therefore,

$\tan \theta =\frac{AB}{BC}\Rightarrow \tan {45}^0=\frac{50\sqrt{3}}{x}\Rightarrow 1=\frac{50\sqrt{3}}{x}(\because \tan {45}^0=1)\Rightarrow x=50\sqrt{3}$

Hence, the distance of the object from the building is $50\sqrt{3}$ m.