The diagonal of a TV is 28 inches long. Assuming that this diagonal forms a pair of 30-60-90 right triangles, what are the exact length and width of the TV?

A. [tex]\(56 \sqrt{2}\)[/tex] inches by [tex]\(56 \sqrt{2}\)[/tex] inches
B. 56 inches by [tex]\(56 \sqrt{3}\)[/tex] inches
C. 14 inches by [tex]\(14 \sqrt{3}\)[/tex] inches
D. [tex]\(14 \sqrt{2}\)[/tex] inches by [tex]\(14 \sqrt{2}\)[/tex] inches



Answer :

To determine the length and width of the TV given that the diagonal is 28 inches and it forms a pair of [tex]\(30^\circ-60^\circ-90^\circ\)[/tex] right triangles, we need to understand the properties and side length ratios of a [tex]\(30^\circ-60^\circ-90^\circ\)[/tex] triangle.

Properties of a [tex]\(30^\circ-60^\circ-90^\circ\)[/tex] Triangle:

In a [tex]\(30^\circ-60^\circ-90^\circ\)[/tex] triangle:
- The side opposite the [tex]\(30^\circ\)[/tex] angle (shorter leg) is [tex]\(x\)[/tex].
- The side opposite the [tex]\(60^\circ\)[/tex] angle (longer leg) is [tex]\(x\sqrt{3}\)[/tex].
- The side opposite the [tex]\(90^\circ\)[/tex] angle (hypotenuse) is [tex]\(2x\)[/tex].

Given the diagonal of the TV is 28 inches, it acts as the hypotenuse of the combined [tex]\(30^\circ-60^\circ-90^\circ\)[/tex] triangles.

Step-by-Step Solution:

1. Since the hypotenuse (diagonal) of each of these [tex]\(30^\circ-60^\circ-90^\circ\)[/tex] triangles is the same, we can isolate one triangle for simplicity. The diagonal forms the hypotenuse of the single mentioned right triangle.

2. Set the hypotenuse [tex]\(2x\)[/tex] equal to 28 inches:
[tex]\[ 2x = 28 \][/tex]

3. Solve for [tex]\(x\)[/tex] (the shorter leg of the triangle):
[tex]\[ x = \frac{28}{2} = 14 \text{ inches} \][/tex]

4. Determine the longer leg (side opposite the [tex]\(60^\circ\)[/tex] angle), which is [tex]\(x\sqrt{3}\)[/tex]:
[tex]\[ \text{Longer leg} = 14\sqrt{3} \text{ inches} \][/tex]

Thus, the exact dimensions of the TV are:
- Width (shorter leg) = 14 inches
- Length (longer leg) = [tex]\(14 \sqrt{3}\)[/tex] inches

Answer:
C. 14 inches by [tex]\(14 \sqrt{3}\)[/tex] inches