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.

check the picture below

and recall that

[tex]\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[/tex]

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lublana lublana · Answer 2 · 2019-07-08T13:13:53+02:00

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

[tex]\frac{DE}{DC}=cos75^{\circ}[/tex]

[tex]\frac{x}{8}=0.2588[/tex]

[tex]x= 0.2588\times8[/tex]

x= 2.070

DE=x=2.0 feet( approximately)

[tex]\frac{CE}{DC}=sin75^{\circ}[/tex]

[tex]\frac{y}{8}=0.9659[/tex]

[tex]y=8\times0.9659[/tex]

y=7.7274

y=7.73( approximately)

CE=7.73 feet

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

In triangle AEC

Usin pythogorous theorem

[tex](CE)^2+(EA)^2=(AC)^2[/tex]

[tex](7.73)^2+(20)^2=(AC)^2[/tex]

[tex](AC)^2= 59.7529+400[/tex]

[tex]AC=\sqrt{459.7529}[/tex]

AC=21.44 feet (approximately)

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