Question

A tower subtends angles $\alpha ,2\alpha ,3\alpha$ at $A,B,C$ all lying on the horizontal line through the foot of the tower, then $\frac{AB}{BC}$ is equal to

• A. $sin2\alpha$
• B. $1+2\cos 2\alpha$
• C. $2+2\cos 2\alpha$
• D. $\frac{\sin 2\alpha }{\sin \alpha }$

Answer 1

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

In any $\Delta PQR$,
Area $\Delta PQR=\frac{1}{2}pq\sin R=\frac{1}{2}qr\sin P=\frac{1}{2}pr\sin Q$
$\Rightarrow \frac{p}{\sin P}=\frac{q}{\sin Q}=\frac{r}{\sin R}$ (This is also known as Sine Rule)
In the given figure, let us consider $\Delta BCE$ ,

In $\Delta BCE$ ,

By exterior angle property,

$\angle AEB=\angle BEC=\alpha$
$\frac{BC}{BE}=\frac{\sin \alpha }{\sin (\pi -3\alpha )}$ ...(using Sine rule)
$AB=BE$ (Since $\Delta ABE$ is an isosceles triangle)

(base angles are equal)
$\Rightarrow \frac{AB}{BC}=\frac{\sin 3\alpha }{\sin \alpha }$
$\Rightarrow \frac{AB}{BC}=3-4{\sin }^2\alpha =1+2\cos 2\alpha$