Question

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 $30^{∘}$, $45^{∘}$ and $60^{∘}$ respectively. The ratio of $AB$ to $BC$ is

• A. $\sqrt{3}:1$
• B. $\sqrt{3}:2$
• C. $1:2$
• D. $2:\sqrt{3}$

Answer 1

Answer: (A)

By sine law, in $â–³ABQ$
$\frac{\sin 15^{∘}}{AB}=\frac{\sin 30^{∘}}{BQ}$

$⇒BQ=\frac{\sin 30^{∘}â‹\dots AB}{\sin 15^{∘}}$
By sine law, in $â–³BCQ$
$\frac{\sin 120^{∘}}{BQ}=\frac{\sin 15^{∘}}{BC}$
$⇒BQ=\frac{BC\sin 120^{∘}}{\sin 15^{∘}}$... (ii)
From Eqs. (i) and (ii), we get
$\frac{\sin 30^{∘}â‹\dots AB}{\sin 15^{∘}}=\frac{BCâ‹\dots \sin 120^{∘}}{\sin 15^{∘}}$
$⇒\frac{AB}{BC}=\frac{\sin 120^{∘}}{\sin 30^{∘}}=\frac{\cos 30^{∘}}{\sin 30^{∘}}=\frac{\sqrt{3}/2}{1/2}$
$AB:BC=\sqrt{3}:1$