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Homework Help: Algebra 2 & Trigonometry

  1. Aug 1, 2010 #1
    1. The problem statement, all variables and given/known data

    A parallelogram's two sides are 24 feet amd 30 feet. The measure of the angle between these sides is 57 degrees. Find the area of the parallelogram, to the nearest square foot.

    2. Relevant equations

    area = 1/2 (a) (b) Sin C

    3. The attempt at a solution

    I'm thinking if there's a 90 degree angle and then the other angle would be 33 degrees. But I'm just assuming this but when I tried it, this did not work. I'm a little troubled by this problem, thanks a lot for your responses.
     
  2. jcsd
  3. Aug 1, 2010 #2

    vela

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    This formula isn't for a parallelogram.
    How did you come up with 33 degrees? If the angle was 90 degrees, wouldn't a parallelogram be a rectangle?
     
  4. Aug 1, 2010 #3
    I came up with 33 degrees by adding 90 and 57 and i got 147 and subtracted 180 by 146and got 33 degrees. But when looking at the properties of a parallelogram, there's no right angles. There are congruent sides and angles but no right angle.
     
  5. Aug 1, 2010 #4
    Okay, then the area of the parallelogram is a = b * h. But then which one would be the base? 24 feet or 30 feet?
     
  6. Aug 1, 2010 #5

    vela

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    Choose one to be the base, then calculate the height using the other one and the angle. It might help you to sketch the parallelogram and draw in the height so you can see how to find its length from the length of the non-base side and the angle.

    You'll find it doesn't make a difference in calculating the area which side you decide is the base.
     
  7. Aug 1, 2010 #6
    How do you calculate the height when you have the base given?
     
  8. Aug 1, 2010 #7

    Mark44

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    Edit: Use correct angle measure.
    If you have a right triangle whose hypotenuse was 24' and the angle between the base and hypotenuse was [STRIKE]33 deg.[/STRIKE] 57 deg., can you find the altitude of this triangle? That would give you the height of your parallelogram.
     
    Last edited: Aug 2, 2010
  9. Aug 2, 2010 #8

    HallsofIvy

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    Almost. The area of a parallelogram with side lengths a and b and angle C between them is just
    ab sin(C)- no "1/2".

    You are told that the angle between the sides is 57 degrees: the area is (24)(30) cos(57). You only need one angle, why look for another?

    Except in the special case of a rectangle, there is no 90 degree angle in a parallelogram. If one angle is 57 degrees, the other angle is 180- 57= 123 degrees. But cos(123)= cos(180- 57)= -sin(57) and the area formula is really an absolute value so it doesn't matter which you use.
     
    Last edited by a moderator: Aug 8, 2010
  10. Aug 7, 2010 #9
    Okay, this is one of the questions from algebra and trigonometry regent in june. It's number 34. I'm pretty sure it doesn't have a 90 degree angle. They all ready give you the answer which is 604 but i dont understand on how to get to that answer.
     
  11. Aug 7, 2010 #10

    Mark44

    Staff: Mentor

    I have to disagree with part of HallsofIvy's answer. The area of the parallelogram is 30*24*sin(57deg), not 30*24*cos(57deg).

    Assuming the long sides of the parallelogram are horizontal, and two of the angles are 57deg, the altitude is 24 sin(57deg).
     
  12. Aug 7, 2010 #11
    Thanks a lot for this everybody and especially @ Mark44
     
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