A parallelogram has side lengths of 13 and 17 and an angle that measures [tex][tex]$64^{\circ}$[/tex][/tex].

Using the Law of Cosines: [tex][tex]$a^2 = b^2 + c^2 - 2bc \cos(A)$[/tex][/tex], what is [tex][tex]$x$[/tex][/tex], the length of the diagonal, to the nearest whole number?

A. 16
B. 18
C. 19
D. 21



Answer :

To solve for the length of the diagonal [tex]\(x\)[/tex] in the parallelogram with given side lengths of 13 and 17 and an angle of [tex]\(64^{\circ}\)[/tex], we can use the Law of Cosines, which is:

[tex]\[a^2 = b^2 + c^2 - 2bc \cos(A)\][/tex]

where:
- [tex]\(a\)[/tex] is the diagonal length we want to find (denoted as [tex]\(x\)[/tex] in the problem),
- [tex]\(b\)[/tex] is the first side length (13),
- [tex]\(c\)[/tex] is the second side length (17),
- [tex]\(A\)[/tex] is the given angle (64 degrees).

Here’s the step-by-step solution:

1. Convert the angle from degrees to radians because trigonometric functions typically use radians. However, we will assume that conversion happens internally during the solving process, directly applying cosine function to the angle.

2. Apply the Law of Cosines to find the square of the diagonal length:
[tex]\[ x^2 = 13^2 + 17^2 - 2 \cdot 13 \cdot 17 \cdot \cos(64^\circ) \][/tex]

3. Compute the individual terms:
- [tex]\(13^2 = 169\)[/tex]
- [tex]\(17^2 = 289\)[/tex]
- [tex]\(\cos(64^\circ) \approx 0.4383711467890774\)[/tex] (a specific approximation may be involved)

4. Calculate the product of the sides and cosine term:
[tex]\[ 2 \cdot 13 \cdot 17 \cdot \cos(64^\circ) = 2 \cdot 13 \cdot 17 \cdot 0.4383711467890774 \approx 194.068 \][/tex]

5. Substitute the values into the Law of Cosines equation:
[tex]\[ x^2 = 169 + 289 - 194.068 \approx 263.932 \][/tex]

6. Find the square root of the result to get the length of the diagonal:
[tex]\[ x = \sqrt{263.932} \approx 16.24 \][/tex]

7. Round to the nearest whole number, which is:
[tex]\[ x \approx 16 \][/tex]

Thus, the length of the diagonal [tex]\(x\)[/tex] to the nearest whole number is [tex]\(16\)[/tex]. The correct answer is:
[tex]\[ \boxed{16} \][/tex]