Ch. 7 - Applications of Trigonometry and Vectors

Lial12th EditionTrigonometryISBN: 9780136552161Not the one you use?Change textbook

Lial 12th Edition

Ch. 7 - Applications of Trigonometry and Vectors

Problem 48

Chapter 8, Problem 48

Starting at point X, a ship sails 15.5 km on a bearing of 200°, then turns and sails 2.4 km on a bearing of 320°. Find the distance of the ship from point X.

Verified step by step guidance

Understand the problem: The ship starts at point X, sails 15.5 km on a bearing of 200°, then sails 2.4 km on a bearing of 320°. We need to find the straight-line distance from the final position back to point X.

Convert the bearings into standard angles relative to the positive x-axis (East). Bearings are measured clockwise from North (0°). So, for a bearing \( \theta_b \), the angle from the positive x-axis is \( \theta = 90° - \theta_b \). Calculate the angles for both legs:

First leg angle: \( \theta_1 = 90° - 200° = -110° \) (which can be interpreted as 250° in standard position).

Second leg angle: \( \theta_2 = 90° - 320° = -230° \) (which can be interpreted as 130° in standard position).

Find the coordinates of the ship after each leg by converting polar coordinates to Cartesian coordinates using \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Then, sum the vectors to find the final position relative to point X. Finally, use the distance formula \( d = \sqrt{(x_{final})^2 + (y_{final})^2} \) to find the distance from point X.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Bearings and Direction

Bearings are angles measured clockwise from the north direction to indicate direction. A bearing of 200° means the ship is sailing 20° west of due south, while 320° means 40° west of due north. Understanding bearings helps translate directional information into angles for calculations.

Vector Representation of Displacement

Each leg of the ship's journey can be represented as a vector with magnitude (distance sailed) and direction (bearing). Converting these vectors into components along the north-south and east-west axes allows for the calculation of the resultant displacement from the starting point.

Distance Calculation Using the Pythagorean Theorem

After finding the resultant vector components, the distance from the starting point is the magnitude of this vector. The Pythagorean theorem is used to calculate this distance by taking the square root of the sum of the squares of the north-south and east-west components.

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Related Practice

Textbook Question

Starting at point A, a ship sails 18.5 km on a bearing of 189°, then turns and sails 47.8 km on a bearing of 317°. Find the distance of the ship from point A.

Textbook Question

A plane has an airspeed of 520 mph. The pilot wishes to fly on a bearing of 310°. A wind of 37 mph is blowing from a bearing of 212°. In what direction should the pilot fly, and what will be her ground speed?

Textbook Question

Find the area of each triangle using the formula 𝓐 = ½ bh, and then verify that the formula 𝓐 = ½ ab sin C gives the same result.

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Textbook Question

One boat pulls a barge with a force of 100 newtons. Another boat pulls the barge at an angle of 45° to the first force, with a force of 200 newtons. Find the resultant force acting on the barge, to the nearest unit, and the angle between the resultant and the first boat, to the nearest tenth.

Textbook Question

Write each vector in the form a i + b j.

〈2, 0〉

Textbook Question

Solve each problem. See Examples 5 and 6.

Distance and Direction of a Motorboat A motorboat sets out in the direction N 80° 00′ E. The speed of the boat in still water is 20.0 mph. If the current is flowing directly south, and the actual direction of the motorboat is due east, find the speed of the current and the actual speed of the motorboat.

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