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Chapter 4: Problem 44

A 200 -foot cliff drops vertically into the ocean. If the angle of elevation from a ship to the top of the cliff is \(22.3^{\circ},\) how far off shore, to the nearest foot, is the ship?

Short Answer

Expert verified
The ship is about \(503\) feet offshore.

Step by step solution

01

Identifying known values

In this situation, the height of the cliff which is \(200\) feet will serve as the opposite side (O) of the right triangle. The angle of elevation of \(22.3^{\circ}\) is also given.
02

Recalling the tangent formula

In any right triangle, the tangent of an angle is given by the ratio of the length of the side opposite the angle to the length of the adjacent side. It can be represented as \(\tan (\theta) = \frac{O}{A}\) where \(\theta\) represents the angle, \(O\) represents the length of the side opposite and \(A\) represents the length of the adjacent side.
03

Using the formula to calculate the distance of the ship

An adaptation of the formula replacing \(O\) and \(\theta\) by their known values gives \(\tan (22.3^{\circ}) = \frac{200}{A}\). To isolate \(A\) (which represents the distance of the ship from the shore), this equation can be reformed as \(A = \frac{200}{\tan (22.3^{\circ})}\).
04

Calculating the distance

Performing the calculation results in \(A \approx 503\) feet. This represents the distance of the ship from the base of the cliff and should be rounded to the nearest foot.

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