hill that has a 37.7 % grade is one that rises 37.7 m vertically for every 100.0 m of distance in the horizontal direction. At what angle is such a hill inclined above the horizontal?
According to the question:the hill rises at 37.7m to the vertical [y-axis] and 100m to the horizontal[x-axis] in the process a right-angle triangle is obtain and an angle of 90 is also obtain. 90 is the angle inclined at the horizontal
Grade = 100%*sin A = 37.7%
sin A = 37.7%/100% = 0.377
A = 22.1o
To climb a hill, the angle of inclination must be less than 90o.
To find the angle at which such a hill is inclined above the horizontal, you can use the trigonometric function called inverse tangent (also known as arctangent). The inverse tangent function will give you the angle when you know the ratio of the rise (vertical distance) to the run (horizontal distance).
Step 1: Convert the grade percentage to its decimal form.
To do this, divide the given grade percentage by 100.
37.7% ÷ 100 = 0.377
Step 2: Find the arctangent of the decimal form.
Using a calculator or a computer program with an arctangent function, find the arctangent of 0.377.
arctan(0.377) = 21.8 degrees
Therefore, the hill is inclined above the horizontal at an angle of approximately 21.8 degrees.