Answer :
Sure, let's solve this step-by-step using the properties of a parallelogram.
Given:
- Diagonal 1 ([tex]\(d_1\)[/tex]) = 10 feet
- Diagonal 2 ([tex]\(d_2\)[/tex]) = 16 feet
- Angle between the diagonals ([tex]\(\theta\)[/tex]) = 28°
The diagonals of a parallelogram bisect each other. Therefore, each diagonal is divided into two equal halves where they intersect. So, each half of diagonal 1 is [tex]\( \frac{10}{2} = 5 \)[/tex] feet and each half of diagonal 2 is [tex]\( \frac{16}{2} = 8 \)[/tex] feet.
Using the law of cosines for the triangle formed by half-diagonals and one side of the parallelogram:
1. Let's denote the halves of the diagonals as [tex]\(d_{1/2}\)[/tex] and [tex]\(d_{2/2}\)[/tex]:
- [tex]\(d_{1/2} = 5\)[/tex] feet
- [tex]\(d_{2/2} = 8\)[/tex] feet
2. The law of cosines in this context can be written as:
[tex]\[ a^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 + 2 \left(\frac{d_1}{2}\right) \left(\frac{d_2}{2}\right) \cos(\theta) \][/tex]
[tex]\[ b^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 - 2 \left(\frac{d_1}{2}\right) \left(\frac{d_2}{2}\right) \cos(\theta) \][/tex]
Since we need the lengths of the sides, which we call [tex]\(s_1\)[/tex] and [tex]\(s_2\)[/tex]:
[tex]\[ s_1 = \sqrt{\left(\frac{10}{2}\right)^2 + \left(\frac{16}{2}\right)^2 + 2 \left(\frac{10}{2}\right) \left(\frac{16}{2}\right) \cos(28^\circ)} \][/tex]
[tex]\[ s_2 = \sqrt{\left(\frac{10}{2}\right)^2 + \left(\frac{16}{2}\right)^2 - 2 \left(\frac{10}{2}\right) \left(\frac{16}{2}\right) \cos(28^\circ)} \][/tex]
Converting the angle to radians:
[tex]\[ \theta = 28^\circ = 28 \times \frac{\pi}{180} \approx 0.4887 \text{ radians} \][/tex]
Substitute in the values and solve:
1. Calculation for [tex]\(s_1\)[/tex]:
[tex]\[ s_1 = \sqrt{5^2 + 8^2 + 2 \cdot 5 \cdot 8 \cdot \cos(0.4887)} \][/tex]
[tex]\[ s_1 = \sqrt{25 + 64 + 80 \cdot \cos(0.4887)} \][/tex]
[tex]\[ s_1 = \sqrt{25 + 64 + 80 \cdot 0.883} \][/tex]
[tex]\[ s_1 = \sqrt{25 + 64 + 70.64} \][/tex]
[tex]\[ s_1 = \sqrt{159.64} \approx 12.63 \text{ feet} \][/tex]
2. Calculation for [tex]\(s_2\)[/tex]:
[tex]\[ s_2 = \sqrt{5^2 + 8^2 - 2 \cdot 5 \cdot 8 \cdot \cos(0.4887)} \][/tex]
[tex]\[ s_2 = \sqrt{25 + 64 - 80 \cdot \cos(0.4887)} \][/tex]
[tex]\[ s_2 = \sqrt{25 + 64 - 70.64} \][/tex]
[tex]\[ s_2 = \sqrt{18.36} \approx 4.29 \text{ feet} \][/tex]
Therefore, the lengths of the sides of the parallelogram are approximately [tex]\( \text{12.63 feet} \)[/tex] and [tex]\( \text{4.29 feet} \)[/tex].
Given:
- Diagonal 1 ([tex]\(d_1\)[/tex]) = 10 feet
- Diagonal 2 ([tex]\(d_2\)[/tex]) = 16 feet
- Angle between the diagonals ([tex]\(\theta\)[/tex]) = 28°
The diagonals of a parallelogram bisect each other. Therefore, each diagonal is divided into two equal halves where they intersect. So, each half of diagonal 1 is [tex]\( \frac{10}{2} = 5 \)[/tex] feet and each half of diagonal 2 is [tex]\( \frac{16}{2} = 8 \)[/tex] feet.
Using the law of cosines for the triangle formed by half-diagonals and one side of the parallelogram:
1. Let's denote the halves of the diagonals as [tex]\(d_{1/2}\)[/tex] and [tex]\(d_{2/2}\)[/tex]:
- [tex]\(d_{1/2} = 5\)[/tex] feet
- [tex]\(d_{2/2} = 8\)[/tex] feet
2. The law of cosines in this context can be written as:
[tex]\[ a^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 + 2 \left(\frac{d_1}{2}\right) \left(\frac{d_2}{2}\right) \cos(\theta) \][/tex]
[tex]\[ b^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 - 2 \left(\frac{d_1}{2}\right) \left(\frac{d_2}{2}\right) \cos(\theta) \][/tex]
Since we need the lengths of the sides, which we call [tex]\(s_1\)[/tex] and [tex]\(s_2\)[/tex]:
[tex]\[ s_1 = \sqrt{\left(\frac{10}{2}\right)^2 + \left(\frac{16}{2}\right)^2 + 2 \left(\frac{10}{2}\right) \left(\frac{16}{2}\right) \cos(28^\circ)} \][/tex]
[tex]\[ s_2 = \sqrt{\left(\frac{10}{2}\right)^2 + \left(\frac{16}{2}\right)^2 - 2 \left(\frac{10}{2}\right) \left(\frac{16}{2}\right) \cos(28^\circ)} \][/tex]
Converting the angle to radians:
[tex]\[ \theta = 28^\circ = 28 \times \frac{\pi}{180} \approx 0.4887 \text{ radians} \][/tex]
Substitute in the values and solve:
1. Calculation for [tex]\(s_1\)[/tex]:
[tex]\[ s_1 = \sqrt{5^2 + 8^2 + 2 \cdot 5 \cdot 8 \cdot \cos(0.4887)} \][/tex]
[tex]\[ s_1 = \sqrt{25 + 64 + 80 \cdot \cos(0.4887)} \][/tex]
[tex]\[ s_1 = \sqrt{25 + 64 + 80 \cdot 0.883} \][/tex]
[tex]\[ s_1 = \sqrt{25 + 64 + 70.64} \][/tex]
[tex]\[ s_1 = \sqrt{159.64} \approx 12.63 \text{ feet} \][/tex]
2. Calculation for [tex]\(s_2\)[/tex]:
[tex]\[ s_2 = \sqrt{5^2 + 8^2 - 2 \cdot 5 \cdot 8 \cdot \cos(0.4887)} \][/tex]
[tex]\[ s_2 = \sqrt{25 + 64 - 80 \cdot \cos(0.4887)} \][/tex]
[tex]\[ s_2 = \sqrt{25 + 64 - 70.64} \][/tex]
[tex]\[ s_2 = \sqrt{18.36} \approx 4.29 \text{ feet} \][/tex]
Therefore, the lengths of the sides of the parallelogram are approximately [tex]\( \text{12.63 feet} \)[/tex] and [tex]\( \text{4.29 feet} \)[/tex].
