anonymous
  • anonymous
A parallelogram has and area of 300 cm squared, the side lengths are 90 cm and 30 cm. What are the measures of the angle?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1362374611174:dw| Can you express the area of the parallelogram in terms of these values. Also the measure of the angle based on trig identities
anonymous
  • anonymous
I dont know what to do with it at all :S
anonymous
  • anonymous
That was all the given information, how an i supposed to find any angles?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Well what is the measure of the angle marked above in terms of the values and variables. Use a rig identity
anonymous
  • anonymous
How tho, i dont understand
anonymous
  • anonymous
theres not enough information given
anonymous
  • anonymous
Can you express the area of the parallelogram in terms of the given values and the variable h we made up to represent the height?
anonymous
  • anonymous
a= 90 x 30
anonymous
  • anonymous
The area of a parallelogram is not side * size. It's base * height You can imagine splitting it into two triangles across the diagonal
anonymous
  • anonymous
Yeah but there's no height given
anonymous
  • anonymous
That's why we are using a variable that we'll solve for. Express the area in terms of the given information and the variable h
anonymous
  • anonymous
We know the length is 90 and width is 30. We need to find the measures of the angles. We also know that the area is 300. Area of parallelogram = L x h (H is height, not the side length of 30) We know Area is 300 and L = 90, but we don't know the height. So we can solve for h. 300 = 90h h = 300/90 = 10/3 Now that we know the height, we can figure out the angles. If you look at Xavier's diagram and the part that he labelled as "angle", that angle is also part of a triangle. That triangle has the height 10/3, same as the parallelogram, and a hypotenuse of 30, which is the other side length of the parallegram. 10/3 is the side opposite to the angle, and 30 is the hypotenuse. From SOH CAH TOA, we can use sine to figure out the angle. @burhan101 sin(x) = (10/3) / (30) = 10/90 = 1/9 We can use the inverse sine function to find angle x. x = sin^-1(1/9) = 6.38 Another thing we should is that one obtuse and one acute angle of a parallelogram are supplementary, meaning they add up to 180. If we know angle x,which is the acute angle is 6.38 degrees, we can subtract this from 180 to find the other angle, let's call it y : 6.38 + y = 180 --> y = 180 - 6.38 = 173.62 Therefore, the two angles x and y are x = 6.38 degrees, y = 173.62 degrees
anonymous
  • anonymous
Area of a parallelogram is also magnitude of vector a and vector b
RadEn
  • RadEn
u can use the law of sin : |dw:1362375710504:dw|
RadEn
  • RadEn
solve for theta after u got the value of sin(theta)
RadEn
  • RadEn
then for the other angle, just use the property of "consecutive interior angle" they sum 180 degrees, let the other angle is alpha : |dw:1362376183926:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.