Question

1.) A 2600 lb. car travelling downhill has a grade resistance of -130 lbs. Find the angle of the grade to the nearest tenth of a degree.
2.) A car travelling on a 2.7 degree uphill grade has a grade resistance of 120 lbs. Determine the weight of the car to the nearest hundred poulds.

Answer 1

Um i think...well i dont know acually

Answer 2

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(or F) = Wsin∅

--> 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)

1. W: 2600lbs       R: -130lbs      ∅: ??
2. (plug in) -->   -130lbs = 2600lbs(sin∅)
3. ÷ both sides by 2600lbs
4. -130lbs÷2600lbs (lbs cancel) = -0.05
5. -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$^{-1}$(-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 '_^