Question

If the angles of elevation of the top of a tower from three collinear points $A,B$ and $C$, on a line leading to the foot of the tower, are ${30}^o,{45}^o$ and ${60}^o$ respectively, then the ratio, $AB:BC$, is

• A. $\sqrt{3}:1$
• B. $\sqrt{3}:\sqrt{2}$
• C. $1:\sqrt{3}$
• D. $2:3$

Answer 1

The correct option is A $\sqrt{3}:1$

In $△PQA$,

$\tan 30=\frac{PQ}{AQ}$

$\Rightarrow AQ=\sqrt{3}h.....\left(1\right)$

In $△PQB$,

$\tan 45=\frac{PQ}{BQ}$

$\Rightarrow BQ=h.....\left(2\right)$

In $△PQC$,

$\tan 60=\frac{PQ}{CQ}$

$\Rightarrow CQ=\frac{h}{\sqrt{3}}$

Now,

$\frac{AB}{BC}=\frac{AQ-BQ}{BQ-CQ}=\frac{\sqrt{3}-1}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}}{1}$