Answer:
Step-by-step explanation:
Using the formula for Grade Resistance/Force in this case...
!!AND MAKING SURE YOUR CALCULATOR IS IN DEGREE MODE!!
--> R/F: the Grade Resistance
--> W: Weight of the Vehicle
--> ∅: Angle of the Hill
Because the Grade Resistance in problem 1) is (-)130lbs... we know that the car is traveling DOWNhill. (If it were simply 130lbs, the car would be traveling up the hill)
- W: 2600lbs R: -130lbs ∅: ??
- (plug in) --> -130lbs = 2600lbs(sin∅)
- ÷ both sides by 2600lbs
- -130lbs÷2600lbs (lbs cancel) = -0.05
- -0.05 = sin∅
- YOU ARE NOT DONE! DO NOT BE FOOLED when you get to this point.
- we have just found SIN∅.... but we were asked to find ∅
to do this, we must take the INVERSE of sin.... (because we cannot simply divide both sides by 'sin' by itself... we must instead...
---> sin[tex]^{-1}[/tex](-0.05) = ?
∅ ≅ -2.866 ..... (you can then divide 866 by 60' if necessary, to find degree and minute answer)...
∅ ≅ -2° 14'
Hope this will help you and others with the 2nd question as well! ^_^
DONT GIVE UP! You got this '_^
