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|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
I dont know what to do with it at all :S
That was all the given information, how an i supposed to find any angles?
Well what is the measure of the angle marked above in terms of the values and variables. Use a rig identity
How tho, i dont understand
theres not enough information given
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?
a= 90 x 30
The area of a parallelogram is not side * size. It's base * height You can imagine splitting it into two triangles across the diagonal
Yeah but there's no height given
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
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
Area of a parallelogram is also magnitude of vector a and vector b
u can use the law of sin : |dw:1362375710504:dw|
solve for theta after u got the value of sin(theta)
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|