Question 1025987: I have a parallelogram, and I need to find the angles.
The measurements I have are that the sides are 12 and 22, and the longer diagonal is 28.
The problem is that the calculated angles I arrived at don t make sense.
I arrive at a degree of 107.2 for part of an angle,with the complementary angle being 72.8. However, using the law of alternate interior angles, the two triangles I have in the parallelogram (after cutting it in half would each have an angle being 107.2, and 72.8. The problem with that naturally is that the 3rd angle in each triangle would have to be 0...
Found 2 solutions by Alan3354, rothauserc:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! I have a parallelogram, and I need to find the angles.
The measurements I have are that the sides are 12 and 22, and the longer diagonal is 28.
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Call the angle opposite the diagonal D.
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Use the Cosine Law:
d^2 = 12^2 + 22^2 - 2*12*22*cos(D) = 28^2
cos(D) = 156/528
D = 107.2 degs
--> 2 angles of 107.2 & 2 angles of 72.8 for a sum of 360 degs.
One angle of each triangle is 107.2. The second angle of each is not 72.8 degs.
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28/sin(D) = 12/sin(A)
sin(A) = 12*sin(D)/28 =~ 0.409
A =~ 24.2 degs, not 72.8
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! We have parallelogram ABCD with interior angles a, b, c, d
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Use law of cosines to find angle b with AC = 28
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28^2 = 22^2 + 12^2 - (2*22*12)cos(b)
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cos(b) = −0.295454545
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cos^(-1) (−0.295454545) = 107.184795874
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angle b is approx 107.2 degrees
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therefore,
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angle c is 72.8 because angle b and angle c are supplementary
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angle d = angle b because alternate angles are congruent
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angle c = angle a becuase alternate angles are congruent
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each diagonal partitions the parallelogram into congruent triangles
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note that unless the parallelogram is a rhombus, the diagonals do not bisect the interior angles
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