A guard observes an enemy boat, from an observation tower a height of 180 m above sea-level, to be at an angle of depression of $30^{\circ}$ . Find the distance from between enemy's boat and foot of tower.

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Solution

Let the distance between the enemy's boat and foot of the tower be $x$ .
$tan 30^{\circ} = \frac{x}{180}$
$\frac{1}{\sqrt{3}} = \frac{x}{180}$
$x = 60 \sqrt{3} m$

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A guard observes an enemy boat, from an observation tower a height of 180 m above sea-level, to be at an angle of depression of 30∘. Find the distance from between enemy's boat and foot of tower.

Let the distance between the enemy's boat and foot of the tower be x.tan30∘=x1801√3=x180x=60√3 m

A guard observes an enemy boat, from an observation tower a height of 180 m above sea-level, to be at an angle of depression of $30^{\circ}$ . Find the distance from between enemy's boat and foot of tower.

Let the distance between the enemy's boat and foot of the tower be $x$ .
$tan 30^{\circ} = \frac{x}{180}$
$\frac{1}{\sqrt{3}} = \frac{x}{180}$
$x = 60 \sqrt{3} m$