Chapter 4, Problem 142CR

To determine

To Find : The angle of the grade and the change in elevation for the car.

Answer to Problem 142CR

The angle of the grade is equal to $6.843°$ and the change in elevation of the car is 0.477 miles .

Explanation of Solution

Given Information :

  

The distance the car travels is 4 miles, that is AC is 4 miles . And for these 4 miles the grade is 12%, that is, $\tan A=12\%$ .

Given that the grade for the distance the car travels is 12% , that is,

  $\tan A=12\%=\frac{12}{100}$

Which implies,

  $A={\tan }^{-1}\frac{12}{100}=6.843°$

According to the information given, the distance travelled by the car is 4 miles .

That is AC = 4 miles .

Let, the change in elevation for the car descending the 4mile stretch be x, that is BC be ‘x’ miles.

Therefore,

  $\sin A=\frac{x}{AC}$

Or,

  $x=\sin A\times AC$

  $x=\sin (6.843°)\times 4$

x = 0.477 miles .

Thus, The angle of the grade is equal to $6.843°$ and the change in elevation of the car is 0.477 miles .