Question

A tower subtends angles $\alpha ,2\alpha$ and $3\alpha$ respectively at points A,B and C, all lying on a horizontal line through the foot of the tower. Find $\frac{AB}{BC}$

• A. $\frac{\sin 3\alpha }{\sin 2\alpha }$
• B. $1+2\cos 2\alpha$
• C. $2+\cos 2\alpha$
• D. $\frac{\sin 2\alpha }{\sin \alpha }$

Answer 1

The correct option is B $1+2\cos 2\alpha$

Now $\angle APB+\angle BAP=\angle PBC$ (exterior angle sum property)

$\Rightarrow \angle APB=\angle PBC-\angle BAP$

Hence, $2\alpha -\alpha =\alpha$

Similarly, $\angle BPC=\alpha$

So,$\angle APB=\angle BAP=\alpha$

$\Rightarrow BP=AB$ ....$\left(1\right)$ (since sides opposite to equal angles are equal)

Using sine formula for $△BCP$ ,

$\frac{BC}{\sin \angle BPC}=\frac{BP}{\sin \angle BCP}$

$\sin ({180}^{\circ }-3\alpha )$ as $\angle PCB+3\alpha ={180}^{\circ }$

$\Rightarrow \frac{BC}{\sin \alpha }=\frac{AB}{\sin 3\alpha }$ since $\sin ({180}^{\circ }-A)=\sin A$

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

$\Rightarrow \frac{AB}{BC}=\frac{3\sin \alpha -4{\sin }^3\alpha }{\sin \alpha }$

$\Rightarrow \frac{AB}{BC}=3-4{\sin }^2\alpha$

$\Rightarrow \frac{AB}{BC}=3-4\left(\frac{1-\cos 2\alpha }{2}\right)$ {since $\cos 2A=1-2{\sin }^{2}A$}

$\Rightarrow \frac{AB}{BC}=3-2+2\cos 2\alpha$

$\therefore \frac{AB}{BC}=1+2\cos 2\alpha$