First slide

Applications of trigonometry

Question

$If the angles of elevations of the top of a tower from three collinear points A, B and C on a$
$line leading to the foot of the tower are 30^{\circ}, 45^{\circ}, and 60^{\circ} respectively, then the ratio$ $A B : B C is$

Moderate

A

$\sqrt{3} : \sqrt{2}$
B

$1 : \sqrt{3}$
C

$2 : 3$
D

$\sqrt{3} : 1$

Solution

$\begin{array}{l} Let ' h' be the height of the tower M N \\ From Δ M C N, \tan 60 = \frac{h}{x} \Rightarrow x = h c o t 60^{°}, a n d f r o m ∆ M B N, \tan 45 = \frac{h}{B N} \Rightarrow h = B N \Rightarrow B C = h - x \Rightarrow B C = h - h c o t 60^{°} \\ From Δ M A N, \tan 30 = \frac{h}{A B + h} \Rightarrow A B + h = h c o t 30^{°} \Rightarrow \\ A B = h \cot 30^{\circ} - h \\ ∴ \frac{A B}{B C} = \frac{h (\cot 30^{\circ} - 1)}{h (1 - \cot 60^{\circ})} \\ = \frac{\sqrt{3} - 1}{1 - \frac{1}{\sqrt{3}}} = \sqrt{3} \\ ∴ A B : B C = \sqrt{3} : 1 \end{array}$

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