A river flows due North, and a tower
stands on its left bank. From a point $A$ upstream and on the
same bank as the tower, the elevation of the tower is 60°,
and from a point $B$ just opposite $A$ on the other bank the
elevation is 45°. If the tower is 360 m high, the breadth of
the river is

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$120 \sqrt{6} m$

$240 / \sqrt{3} m$

$240 \sqrt{3} m$

$240 \sqrt{6} m$

answer is A.

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Detailed Solution

Let $P Q$ be the tower of height 360 $m$ standing
on the left bank of the river subtending an angle 60° at
$A$ on the same bank as the tower. (Fig. 27.30) Then $P A =$
$360 c o t 60 °$

Let $B$ be the point opposite to $A$ on the other bank where the
elevation of the tower is 45°. So that $P B = 360 \cot 45^{\circ} = 360$

If $x$ be the width of the river, then $A B = x$ and from the
right angled triangle $P A B$

$\begin{array}{l} \Rightarrow (P A)^{2} + (A B)^{2} = (P B)^{2} \\ \Rightarrow {(360 \times \frac{1}{\sqrt{3}})}^{2} + x^{2} = (360)^{2} \\ \Rightarrow x = 360 \times \frac{\sqrt{2}}{\sqrt{3}} = 120 \sqrt{6} m . \end{array}$