(Modeling) Grade Resistance Solve each problem. See Example 3. A 3000-lb car traveling uphill has a grade resistance of 150 lb. Find the angle of the grade to the nearest tenth of a degree.

Verified step by step guidance

1. Understand that grade resistance is the component of the car's weight acting parallel to the slope. It can be expressed as \(\text{Grade Resistance} = \text{Weight} \times \sin(\theta)\), where \(\theta\) is the angle of the grade.
2. Identify the given values: Weight \(W = 3000\) lb and Grade Resistance \(R = 150\) lb. We need to find the angle \(\theta\) such that \(R = W \times \sin(\theta)\).
3. Set up the equation using the given values: \(150 = 3000 \times \sin(\theta)\).
4. Solve for \(\sin(\theta)\) by dividing both sides of the equation by 3000: \(\sin(\theta) = \frac{150}{3000}\).
5. Find the angle \(\theta\) by taking the inverse sine (arcsin) of the value found: \(\theta = \arcsin\left(\frac{150}{3000}\right)\). Then, convert the result to degrees and round to the nearest tenth.

Key Concepts

Grade Resistance and Its Relation to Weight

Grade resistance is the component of a vehicle's weight acting parallel to an inclined surface, opposing motion uphill. It depends on the weight of the vehicle and the slope angle, representing the force needed to overcome gravity along the incline.

Trigonometric Relationship Between Forces and Angles

The grade resistance force can be modeled as the product of the vehicle's weight and the sine of the incline angle. This relationship allows us to use the sine function to find the angle when the force and weight are known.

Inverse Sine Function to Find the Angle

To determine the incline angle from the grade resistance and weight, the inverse sine (arcsin) function is used. It calculates the angle whose sine equals the ratio of grade resistance to weight, providing the angle in degrees.