Question

The longer base of an isosceles trapezoid measures 22 ft. The nonparallel sides measure 8 ft, and the base angles measure 75°. Find the length of a diagonal.

Answer 1

Answer:

The length of a diagonal is 21.44 feet ( approximately).

Step-by-step explanation:

Given a trapezoid ABCD

AD= 22 feet

DC=8 feet

AD= DE+EA

Let DE=x and CE= y

In triangle DEC

$\frac{DE}{DC}=cos75^{\circ}$

$\frac{x}{8}=0.2588$

$x= 0.2588\times8$

x= 2.070

DE=x=2.0 feet( approximately)

$\frac{CE}{DC}=sin75^{\circ}$

$\frac{y}{8}=0.9659$

$y=8\times0.9659$

y=7.7274

y=7.73( approximately)

CE=7.73 feet

EA= AD-DE=22-2=20 feet

In triangle AEC

Usin pythogorous theorem

$(CE)^2+(EA)^2=(AC)^2$

$(7.73)^2+(20)^2=(AC)^2$

$(AC)^2= 59.7529+400$

$AC=\sqrt{459.7529}$

AC=21.44 feet (approximately)

Hence, the length of diagonal =21.44 feet ( approximately).

Answer 2

Check the picture below

and recall that
 
$\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\ -------------------------------\\\\ sin(75^o)=\cfrac{y}{8}\implies 8sin(75^o)=y \\\\\\ cos(75^o)=\cfrac{x}{8}\implies 8cos(75^o)=x$