Elementary Geometry for College Students (7th Edition)

Published by Cengage

Chapter 11 - Review Exercises - Page 533: 30

Answer

The boats are $\approx 1412.0$ meters apart.

Work Step by Step

There are two main triangles present in the diagram (Figure 1, see attachment): $\triangle ABC$ $\triangle ACD$ $AB =$ Height of the plane $CD =$ Distance between the boats 1. Find all the angles in $\triangle ABC$ $\angle ABC = 90˚$ (Right angled triangle) $\angle BAC = 90 - 44 = 46˚$ (From the horizontal to vertical line = 90˚) $\angle ACB = 180 - (46 + 90) = 44˚$ (Angles in a triangle add to 180˚) 2. Find all the angles in $\triangle ACD$ $\angle CAD = 44 - 32 = 12˚$ (Values given to find the angle) $\angle ACD = 180 - 44 = 136˚$(Angles on a line add to 180˚) $\angle CDA = 180 - (136 + 12) = 32˚$ (Angles in a triangle add to 180˚) 3. Find the distance of $AC$ using the sine law + calculator $\frac{2500}{sin44} = \frac{AC}{sin90}$ $\frac{2500sin90}{sin44} = AC$ $\frac{2500}{sin44} = AC$ $\frac{2500}{0.69466...} = AC$ $3598.9$ meters = $AC$ 4. Find the distance of $CD$ using Step #3, sine law and your calculator $\frac{3598.9}{sin32} = \frac{CD}{sin12}$ $\frac{3598.9sin12}{sin32} = CD$ $\frac{748.2516...}{sin32} = CD$ $\frac{748.2516...}{0.529919...} = CD$ $1412.0$ meters = $CD$ Therefore the distance between the two boats is $\approx 1412.0$ meters.

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