Question

The size of a television is the length of the diagonal of its screen in inches. The aspect ratio of the screens of older televisions is 4:3, while the aspect ratio of newer wide-screen televisions is 16:9. Find the width and height of an older 35-inch television whose screen has an aspect ratio of 4:3.

Answer 1

Answer:

The width and height of the old 35-inch television are 28 inches and 21 inches, respectively.

Step-by-step explanation:

35-inch television is a television whose screen has an hypotenuse ($l$) of 35 inches and the aspect ratio of 4 : 3 means that 4 inches width per each 3 inches height. And by Pythagorean Theorem:

$r_{l} =\sqrt{r_{w}^{2}+r_{h}^{2}}$

Where:

$r_{l}$ - Hypotenuse rate, dimensionless.

$r_{w}$ - Width rate, dimensionless.

$r_{h}$ - Height rate, dimensionless.

If $r_{w} = 4\,in$ and $r_{h} = 3\,in$, the hypotenuse rate is:

$r_{l} = \sqrt{4^{2}+3^{2}}$

$r_{l} = 5$

The width and height of the old television can be found with the help of trigonometric functions:

Width of 35-inch old television ($w$)

$w = l\cdot \cos \theta$

$w = l\times\left(\frac{r_{w}}{r_{l}} \right)$

Height of 35-inch old television ($h$)

$h = l\cdot \sin \theta$

$h = l\times\left(\frac{r_{h}}{r_{l}} \right)$

Where $\theta$ is the direction of the hypotenuse with respect to width, measured in sexagesimal degrees.

If $r_{w} = 4$, $r_{h} = 3$ and $r_{l} = 5$ and $l = 35\,in$, the width and height of the old 35-inch television:

$w = (35\,in)\times \left(\frac{4}{5} \right)$

$w = 28\,in$

$h =(35\,in)\times \left(\frac{3}{5} \right)$

$h = 21\,in$

The width and height of the old 35-inch television are 28 inches and 21 inches, respectively.