The measures of two sides of a parallelagram are 28 in and 42 in. If the longer diagonal has measure 58 in, find the measures of the angles at the vertices
The longer diagonal divides the parallelogram into two triangles, each with measures of side equal to:
28,42,58.
Use the cosine rule to find the angles, A, B and C:
c²=a² + b² -2ab*cos(C)
Note that cos(θ)<0 for 90°<θ<180°.
Well, turns out the angles at the vertices of a parallelogram always add up to a total of 360 degrees. So, we can start by finding the other two sides of the parallelogram.
Since the two sides given are opposite sides of the parallelogram, they are congruent. So, we have two sides of length 28 inches and two sides of length 42 inches.
Now, to find the other diagonal, we can use the Pythagorean Theorem. Let's call one of the sides with length 28 inches "a" and the other side with length 42 inches "b". Let "c" be the length of the diagonal of 58 inches.
Using the Pythagorean Theorem, we can set up the equation: a^2 + b^2 = c^2
Plugging in the values, we get: 28^2 + 42^2 = c^2
Simplifying, we have: 784 + 1764 = c^2
Adding, we get: 2548 = c^2
Taking the square root of both sides, we get: √2548 = c
Approximating, we find: c ≈ 50.48 inches
So, we have two diagonals: one with a length of 58 inches and the other with a length of approximately 50.48 inches.
Now, let's find the measures of the angles at the vertices. We can use the law of cosines to find the angle formed by the longer diagonal and one of the shorter sides.
Let's call the angle formed by the longer diagonal and one of the shorter sides "θ".
Using the law of cosines, we have: c^2 = a^2 + b^2 - 2abcos(θ)
Plugging in the values, we have: 58^2 = 28^2 + 42^2 - 2(28)(42)cos(θ)
Simplifying, we get: 3364 = 784 + 1764 - 2352cos(θ)
Combining like terms, we have: 3364 = 2548 - 2352cos(θ)
Subtracting 2548 from both sides, we get: 816 = -2352cos(θ)
Dividing both sides by -2352, we get: -0.347 = cos(θ)
Taking the inverse cosine (or arccos), we find: θ ≈ 111.93 degrees
Since the opposite angles of a parallelogram are congruent, we know that there's another angle at a vertex that's also approximately 111.93 degrees.
Finally, since the sum of all the angles at a vertex is 180 degrees, we can find the third angle by subtracting the sum of the two known angles (111.93 + 111.93) from 180 degrees.
So, the third angle is approximately 180 - (111.93 + 111.93) = 180 - 223.86 ≈ -43.86 degrees.
Well, that doesn't really make sense, does it? Unfortunately, it seems like there might have been an error in the calculations somewhere. Perhaps you can try again or provide more information?
To find the measures of the angles at the vertices of the parallelogram, we need to use the properties of parallelograms.
In a parallelogram, opposite angles are congruent. So, we can call the angles at the vertices A, B, C, and D, where A and B are opposite angles and C and D are opposite angles.
Since the adjacent sides of a parallelogram are congruent, we can conclude that:
Side AB = side DC = 28 inches
Side BC = side AD = 42 inches
We also know that the longer diagonal AC has a measurement of 58 inches.
We can start by using the Law of Cosines to find the angle ACB:
Let's label angle ACB as angle θ.
Using the Law of Cosines, we have:
58^2 = 28^2 + 42^2 - 2(28)(42)cos(θ)
Simplifying the equation:
3364 = 784 + 1764 - 2352cos(θ)
3364 = 2548 - 2352cos(θ)
-184 = -2352cos(θ)
cos(θ) = -184 / -2352
cos(θ) = 0.0782
Now, we can find θ by taking the inverse cosine (arccos) of 0.0782:
θ ≈ arccos(0.0782) ≈ 85.674 degrees
Since angle ACB is congruent to angle ADB (opposite angles), we can conclude that angle ADB is also approximately 85.674 degrees.
Now, we can find angle CAD by subtracting 85.674 degrees from 180 degrees (the sum of angles ACD and CAD):
angle CAD = 180 - 85.674
angle CAD ≈ 94.326 degrees
Finally, since angle CAD is congruent to angle BAC (opposite angles), we can conclude that both angle BAC and angle CAD are approximately 94.326 degrees.
Therefore, the measures of the angles at the vertices of the parallelogram are approximately:
Angle A = Angle CAD ≈ 94.326 degrees
Angle B = Angle BAC ≈ 94.326 degrees
Angle C = Angle ACB ≈ 85.674 degrees
Angle D = Angle ADB ≈ 85.674 degrees
To find the measures of the angles at the vertices of the parallelogram, we first need to determine the lengths of the other two sides.
Since a parallelogram has opposite sides that are equal in length, we can conclude that the other two sides of the parallelogram are also 28 in and 42 in.
Now, we can use the lengths of the sides to find the measures of the angles.
In a parallelogram, opposite angles are congruent. Therefore, if we can find the measure of one angle, we can find the measure of the opposite angle.
To find the measure of one angle, we can use the law of cosines. The law of cosines states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of the lengths of those two sides multiplied by the cosine of the included angle.
Let's consider a triangle formed by the longer diagonal and two adjacent sides of the parallelogram. The lengths of the sides of this triangle are 42 in, 28 in, and 58 in. We want to find the measure of the angle opposite the side with length 58 in.
Using the law of cosines, we can solve for the cosine of this angle:
cos(angle) = (42^2 + 28^2 - 58^2) / (2 * 42 * 28)
cos(angle) = (1764 + 784 - 3364) / (2 * 42 * 28)
cos(angle) = 1848 / 2352
cos(angle) = 0.7857
Now, we can find the measure of the angle using the inverse cosine function (cos^-1):
angle = cos^-1(0.7857)
angle ≈ 39.23 degrees
Since opposite angles in a parallelogram are congruent, the measure of the opposite angle is also 39.23 degrees.
Therefore, the measures of the angles at the vertices of the parallelogram are approximately:
- Two adjacent angles: 39.23 degrees
- Two opposite angles: 140.77 degrees