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Q.
Angles of elevation of the top of a tower from three points (collinear) $A, B$ and $ C$ on a road leading to the foot of the tower are $30^\circ$, $45^\circ$ and $60^\circ$ respectively. The ratio of $AB$ to $BC$ is
By sine law, in $\triangle A B Q$
$\frac{\sin 15^{\circ}}{A B}=\frac{\sin 30^{\circ}}{B Q}$
$\Rightarrow \quad B Q=\frac{\sin 30^{\circ} \cdot A B}{\sin 15^{\circ}}$
By sine law, in $\triangle B C Q$
$\frac{\sin 120^{\circ}}{B Q}=\frac{\sin 15^{\circ}}{B C}$
$\Rightarrow B Q=\frac{B C \sin 120^{\circ}}{\sin 15^{\circ}}$... (ii)
From Eqs. (i) and (ii), we get
$\frac{\sin 30^{\circ} \cdot A B}{\sin 15^{\circ}}=\frac{B C \cdot \sin 120^{\circ}}{\sin 15^{\circ}}$
$\Rightarrow \frac{A B}{B C}=\frac{\sin 120^{\circ}}{\sin 30^{\circ}}=\frac{\cos 30^{\circ}}{\sin 30^{\circ}}=\frac{\sqrt{3} / 2}{1 / 2}$
$A B: B C=\sqrt{3}: 1$
