Question

A tower subtends angles $\theta ,2\theta$ and $3\theta$ at three points $A,B,C$ respectively lying on a horizontal line through the foot of tower. Then the ratio $\frac{AB}{BC}$ equals

• A. $\frac{\sin 3\theta }{\sin \theta }$
• B. $\frac{\sin \theta }{\sin 3\theta }$
• C. $\frac{\cos 3\theta }{\cos \theta }$
• D. $\frac{\tan \theta }{\tan 3\theta }$

Answer 1

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


$\tan 3\theta =\frac{h}{QC}\Rightarrow QC=h\cot 3\theta$
Similarly,
$QB=h\cot 2\theta QA=h\cot \theta$

$\frac{AB}{BC}=\frac{QA-QB}{QB-QC}\Rightarrow \frac{AB}{BC}=\frac{\cot \theta -\cot 2\theta }{\cot 2\theta -\cot 3\theta }\Rightarrow \frac{AB}{BC}=\frac{\sin 3\theta }{\sin \theta }$