Question

On the same side of a tower two objects are located. Observed from the top of the tower their angles of depression are ${45}^0$ and ${60}^0$. If the height of the tower is $150m$, the distance between the objects is

• A. $63.5$ $m$
• B. $76.9$ $m$
• C. $86.7$ $m$
• D. $90$ $m$

Answer 1

The correct option is A $63.5$ $m$

Let $AB$ be the tower and $C$ and $D$ be the point of object

Given $AB=150m$ and $\angle ACB={45}^0$ and $\angle ADB={60}^0$

In $\Delta BAD$,

$\tan {60}^0=\frac{AB}{AD}$

$\Rightarrow \sqrt{3}=\frac{150}{AD}$

$\Rightarrow AD=\frac{150}{\sqrt{3}}\text{m}$

In triangle $BAC$,

$\tan {45}^0=\frac{AB}{AC}$

$\Rightarrow 1=\frac{150}{AC}$

$\Rightarrow AC=150\text{m}$

$\therefore CD=AC-AD$

$=150-\frac{150}{\sqrt{3}}$

$=\left[\frac{150(\sqrt{3}-1)}{\sqrt{3}}\right]$

$=50(3-\sqrt{3})$

$=50\times 1.27$

$=63.5$ $\text{m}$