   Chapter 11.CR, Problem 30CR ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
1 views

# In Exercises 21 to 30, use the drawings, where provided, to solve each problem, Angle measures should be found to the nearest degree; lengths should be found to the nearest tenth of a unit.An observer in a plane 2500 m high sights two ships below. The angle of depression to one ship is 32 ° , and the angle of depression to the other ship is 44 ° . How far apart are the ships? To determine

To find:

The distance between the ships. An observer in a plane 2500 m high sights two ships below, the angle of depression to one ship is 32° and for other ship. 44°.

Explanation

Procedure used:

In any right triangle ABC with mA=θ we have the following ratios.

sinθ=OppositeHypotenuse

Calculation:

Given:

Height of the plane =2500 m.

Angle of depression of the ships are 32° and 44°.

Let us structure the given figure as a right triangle as follows.

From the above figure we have

CD=2500 m,

mZDA=32°, and

mZDB=44°

Let ‘x’ be the distance between the ships.

AB=x

BC=y (say), and

CA=BC+AB

CA=x+y`

Also by the property that the alternate interior angles are equal we have

mZDA=mDAC=32° and

mZDB=mDBC=44°

In the above figure we have two right angle triangle BCD and ACD.

Take the BCD and apply the ratio tanθ=OppositeAdjacent

tan44°=CDBC

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