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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 $A B$ to $B C$ is
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Verified Answer
The correct answer is:
$\sqrt{3}: 1$
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 $\triangle B C Q$
$$
\begin{gathered}
\frac{\sin 120^{\circ}}{B Q}=\frac{\sin 15^{\circ}}{B C} \\
\Rightarrow \quad B Q=\frac{B C \sin 120^{\circ}}{\sin 15^{\circ}}
\end{gathered}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
\frac{\sin 30^{\circ} \cdot A B}{\sin 15^{\circ}} &=\frac{B C \cdot \sin 120^{\circ}}{\sin 15^{\circ}} \\
\Rightarrow \quad \frac{A B}{B C}-\frac{\sin 120^{\circ}}{\sin 30^{\circ}} &=\frac{\cos 30^{\circ}}{\sin 30^{\circ}}-\frac{\sqrt{3} / 2}{1 / 2} \\
A B: B C &=\sqrt{3}: 1
\end{aligned}
$$
$$
\frac{\sin 15^{\circ}}{A B}=\frac{\sin 30^{\circ}}{B Q}
$$
$$
\Rightarrow \quad B Q=\frac{\sin 30^{\circ} \cdot A B}{\sin 15^{\circ}}
$$
By sine law, in $\triangle B C Q$
$$
\begin{gathered}
\frac{\sin 120^{\circ}}{B Q}=\frac{\sin 15^{\circ}}{B C} \\
\Rightarrow \quad B Q=\frac{B C \sin 120^{\circ}}{\sin 15^{\circ}}
\end{gathered}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
\frac{\sin 30^{\circ} \cdot A B}{\sin 15^{\circ}} &=\frac{B C \cdot \sin 120^{\circ}}{\sin 15^{\circ}} \\
\Rightarrow \quad \frac{A B}{B C}-\frac{\sin 120^{\circ}}{\sin 30^{\circ}} &=\frac{\cos 30^{\circ}}{\sin 30^{\circ}}-\frac{\sqrt{3} / 2}{1 / 2} \\
A B: B C &=\sqrt{3}: 1
\end{aligned}
$$
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