Question

If a tower subtends angles $\theta$, $2\theta$ and $3\theta$ at three points $A,B$ and $C$ respectively, lying on the same side of a horizontal line through the foot of the tower, $\frac{AB}{BC}$ is equal to?

• A. $\frac{\sin 2\theta }{\sin \theta }$
• B. $\frac{\sin 3\theta }{\sin 2\theta }$
• C. $\frac{\sin 3\theta }{\sin \theta }$
• D. $\frac{\cot \theta -\cot 2\theta }{\cot 2\theta -\cot 3\theta }$

Answer 1

The correct option is C $\frac{\sin 3\theta }{\sin \theta }$

Let $PQ$ be the tower s given $\angle PAQ=\theta$
$\angle PBQ=2\theta ,\angle PCQ=3\theta$
$\therefore \angle BPA=\angle CPB=\theta △PAB$
$\angle PAB=\angle BPA=\theta$
$\therefore AB=BP$ -(1)

In $△PBC$ (By) sine rule

$\Rightarrow \frac{BP}{\sin ({180}^0-3\theta )}=\frac{BC}{\sin \theta }$
$\Rightarrow \frac{AB}{\sin 3\theta }=\frac{BC}{\sin \theta }$ {form(1)}

$\Rightarrow \frac{AB}{BC}=\frac{\sin 3\theta }{\sin \theta }$