Textbook Question
CONCEPT PREVIEW Match the measure of bearing in Column I with the appropriate graph in Column II.
I. 10. N 70° E
II. 1. A. B. C. 2. 3. 4. D. E. F. 5. 6. 7. G. H. 8. 9. I. J.
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Ch. 2 - Acute Angles and Right Triangles
Lial12th EditionTrigonometryISBN: 9780136552161
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Table of contents
- Ch. R - Algebra Review381
- Ch. 1 - Trigonometric Functions188
- Ch. 2 - Acute Angles and Right Triangles168
- Ch. 3 - Radian Measure and The Unit Circle155
- Ch. 4 - Graphs of the Circular Functions100
- Ch. 5 - Trigonometric Identities238
- Ch. 6 - Inverse Circular Functions and Trigonometric Equations158
- Ch. 7 - Applications of Trigonometry and Vectors148
- Ch. 8 - Complex Numbers, Polar Equations, and Parametric Equations15
Chapter 3, Problem 2.3.70
(Modeling) Grade Resistance Solve each problem. See Example 3. Find the grade resistance, to the nearest ten pounds, for a 2400-lb car traveling on a -2.4° downhill grade.
Verified step by step guidance1
Understand that grade resistance is the component of the car's weight acting along the slope due to gravity. It can be found using the formula: \(\text{Grade Resistance} = W \times \sin(\theta)\), where \(W\) is the weight of the car and \(\theta\) is the angle of the slope.
Identify the given values: the weight of the car \(W = 2400\) pounds, and the slope angle \(\theta = -2.4^\circ\). The negative sign indicates a downhill grade.
Calculate the sine of the angle \(\theta\). Since the angle is negative, \(\sin(-2.4^\circ)\) will be negative, reflecting the downhill direction.
Multiply the weight \(W\) by \(\sin(\theta)\) to find the grade resistance: \(2400 \times \sin(-2.4^\circ)\).
Interpret the result: the grade resistance will be negative, indicating it acts in the direction of motion (downhill). Round the absolute value of the result to the nearest ten pounds as requested.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Grade Resistance
Grade resistance is the component of gravitational force acting against the motion of a vehicle on an inclined surface. It depends on the vehicle's weight and the slope angle, calculated as the weight multiplied by the sine of the grade angle. This force affects how much effort is needed to move the vehicle uphill or downhill.
Trigonometric Functions and Angles
Understanding sine and its relation to angles is essential for calculating forces on slopes. The sine of the grade angle gives the ratio of the vertical height change to the hypotenuse (road length), allowing conversion of the slope angle into a force component. Negative angles indicate downhill grades, affecting the direction of the force.
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Force Calculation and Units
Calculating grade resistance requires multiplying the vehicle's weight (in pounds) by the sine of the grade angle to find the force in pounds. Proper rounding, as requested to the nearest ten pounds, ensures practical and usable results. Understanding units and rounding conventions is crucial for accurate and meaningful answers.
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Related Practice
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