Problem 13
Question
A straight road leads from an ocean beach into the nearby hills. The road has a constant upward grade of \(3^{\circ} .\) After taking this road for one mile, how high above sea level (in feet) are you?(GRAPH CAN'T COPY)
Step-by-Step Solution
Verified Answer
Answer: Approximately 277.37 feet.
1Step 1: Convert the distance to feet
Since the distance given is in miles, we'll first convert it to feet. 1 mile is equal to 5280 feet.
2Step 2: Set up the right-angled triangle
In the given scenario, we can visualize a right-angled triangle, where the base is the horizontal distance (1 mile or 5280 feet), the height we need to find out is the opposite side, and the angle is \(3^{\circ}\).
3Step 3: Use the tangent function
Since we have the angle (\(3^{\circ}\)) and want to find the opposite side of the right-angled triangle using the adjacent side (5280 feet), we can use tangent function. Recall that \(\tan(\theta) = \frac{opposite}{adjacent}\). We'll rearrange the formula to solve for the opposite side (height above sea level).
4Step 4: Solve for the opposite side
With the formula from the previous step, we can solve for the opposite side (height above sea level) using the given values.
Opposite = Adjacent \(\times\) \(\tan{(\theta)}\)
Opposite = 5280 \(\times\) \(\tan{(3^{\circ})}\)
After calculating, we'll find the height above sea level:
Opposite (height above sea level) \(\approx\) 277.37 feet.
Hence, after traveling one mile along the straight road with a constant upward grade of \(3^{\circ}\), you are approximately 277.37 feet above sea level.
Key Concepts
Understanding Right-Angled TrianglesThe Tangent Function in TrigonometryCalculating the Grade of a Road
Understanding Right-Angled Triangles
A right-angled triangle is a fundamental shape in geometry where one of its angles measures exactly 90 degrees. This triangle is significant in trigonometry, as it forms the basis for defining trigonometric ratios like sine, cosine, and tangent. In our exercise, we consider the path of a road against the level ground to form a right-angle. Think of it as a triangle sliced vertically out of the hillside where your path goes upwards. The base of this triangle, known as the 'adjacent' side in trigonometry, is the distance covered along the road, while the height or 'opposite' side, is the elevation from the initial point (sea level) after traveling 1 mile uphill. Understanding this triangle's structure helps us use trigonometric functions to solve real-world problems like finding the height of a hill or the grade of a road.
Remember, every time you have a right angle and one other angle and one side of a triangle, you have enough information to find out everything else about that triangle. That is the power of right-angled triangles in trigonometry!
Remember, every time you have a right angle and one other angle and one side of a triangle, you have enough information to find out everything else about that triangle. That is the power of right-angled triangles in trigonometry!
The Tangent Function in Trigonometry
The tangent function is incredibly useful when dealing with right-angled triangles in where we know one acute angle and one side. It relates the angle to the ratio of the opposite side over the adjacent side. In mathematical terms, for an angle \(\theta\), the tangent is denoted as \(\tan(\theta)\) and is calculated by dividing the length of the side opposite \(\theta\) by the length of the side next to \(\theta\).
In our road gradient example, we used the tangent function to find the height above sea level. With the given angle of elevation (\(3^\circ\)), we applied the tangent function as follows: \(\tan(3^\circ) = \frac{opposite}{5280}\). This formula was then rearranged to solve for the opposite side. It's essential to get comfortable with using the tangent function and rearranging equations to solve for the unknown, as this is a common task in trigonometry.
In our road gradient example, we used the tangent function to find the height above sea level. With the given angle of elevation (\(3^\circ\)), we applied the tangent function as follows: \(\tan(3^\circ) = \frac{opposite}{5280}\). This formula was then rearranged to solve for the opposite side. It's essential to get comfortable with using the tangent function and rearranging equations to solve for the unknown, as this is a common task in trigonometry.
Practical Usage:
Understanding how to utilize the tangent function is not just about solving textbook problems; it's a practical skill. Engineers, architects, and even smartphone apps rely on this function to calculate angles and distances.Calculating the Grade of a Road
When we talk about the grade of a road in precalculus and real-world applications, we're referring to the steepness or incline of the road. It is usually expressed as a percentage, angle, or ratio. The road's grade can tell you how steep a hill is. For drivers, a steeper grade can affect vehicle performance and fuel efficiency.
In the given exercise, the road's grade is represented as an angle of ascent, which is \(3^\circ\) in this case. This angle indicates how quickly the road rises compared to the horizontal distance traveled. To calculate the elevation gain - how high the road goes - we can use the tangent function, as previously discussed: by multiplying the road's length in feet by the tangent of the grade angle.
In the given exercise, the road's grade is represented as an angle of ascent, which is \(3^\circ\) in this case. This angle indicates how quickly the road rises compared to the horizontal distance traveled. To calculate the elevation gain - how high the road goes - we can use the tangent function, as previously discussed: by multiplying the road's length in feet by the tangent of the grade angle.
Real-world Implications:
Beyond just measurement, understanding grades is essential for civil engineering and construction, determining whether a road or pathway is suitable for public use, and even for creating pathways accessible to individuals with disabilities.Other exercises in this chapter
Problem 12
Directions: Standard notation for triangle \(A B C\) is used throughout. Use a calculator and round off your answers to one decimal place at the end of the comp
View solution Problem 13
Solve the triangle. The Law of Cosines may be needed. $$a=5, c=12, A=102^{\circ}$$
View solution Problem 13
Directions: Standard notation for triangle \(A B C\) is used throughout. Use a calculator and round off your answers to one decimal place at the end of the comp
View solution Problem 14
Solve the triangle. The Law of Cosines may be needed. $$a=9, b=14, B=95^{\circ}$$
View solution