Question

On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are ${45}^o$ and ${60}^o$. If the height of the tower is $50\sqrt{3}$, then the distance between the objects is

• A. $\frac{50}{\sqrt{3}}$m
• B. $150$m
• C. $50(\sqrt{3}-1)$m
• D. $25(\sqrt{3}-1)$m

Answer 1

The correct option is C $50(\sqrt{3}-1)$m

Let AB be the tower which is $50\sqrt{3}m$ high and C and D be the positions of the two towers such that

$\angle ADB={45}^0$ and $\angle ACB={60}^0$

Now, in triangle ABC,

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

$\Rightarrow$$\sqrt{3}=\frac{50\sqrt{3}}{BC}$

$\Rightarrow$$BC=50m$

Also, in triangle ABD,

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

$\Rightarrow \tan {45}^0=\frac{AB}{BC+CD}$

$\Rightarrow$$1=\frac{50\sqrt{3}}{50+CD}$

$\Rightarrow$$50+CD=50\sqrt{3}$

$\Rightarrow$$CD=50\sqrt{3}-50$

$\Rightarrow$$CD=50(\sqrt{3}-1)m$.

Hence option C is correct.