# How do I find the base angle of an isosceles trapezoid with bases 10 and 18 in length and a leg that is 8 in length?

Dec 14, 2015

$\textcolor{w h i t e}{\times} 60 \textcolor{w h i t e}{x}$Degrees

#### Explanation: $\textcolor{w h i t e}{\times} | E F | = | D C |$
$\textcolor{w h i t e}{\times \times x} = 10 \textcolor{w h i t e}{x} \text{in}$

$\textcolor{w h i t e}{\times} | A D | = | B C | \textcolor{w h i t e}{\times \times \times \times \times x}$(isosceles trapezoid),
$\textcolor{w h i t e}{\times} | D E | = | C F | \textcolor{w h i t e}{\times \times \times \times \times x}$(isosceles trapezoid),
$\textcolor{w h i t e}{\times} m \left(\angle A D E\right) = m \left(\angle B C F\right) \textcolor{w h i t e}{\times x}$(isosceles trapezoid),

$\textcolor{red}{\text{SAS Postulate}}$: Two sides in a triangle have the same length as two sides in the other triangle, and the included angles have the same measure. Therefore

$\textcolor{w h i t e}{\times} \Delta A D E$
and
$\textcolor{w h i t e}{\times} \Delta B C F$
are congruent:
$\textcolor{w h i t e}{\times} \Delta A D E = \Delta B C F$

$\textcolor{w h i t e}{\times} | A B | = 18 \textcolor{w h i t e}{x} \text{in} \textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times}$(base length)
$\implies | A E | + | E F | + | F B | = 18 \textcolor{w h i t e}{x} \text{in} \textcolor{w h i t e}{\times \times \times \times \times x}$(base length)
$\implies | A E | + | E F | + \textcolor{red}{|} A E | = 18 \textcolor{w h i t e}{x} \text{in} \textcolor{w h i t e}{\times \times \times \times \times}$(SAS Postulate)
$\implies | A E | + 10 + | A E | = 18$
$\implies | A E | + 10 + | A E | \textcolor{red}{- 10} = 18 \textcolor{red}{- 10}$
$\implies \textcolor{red}{\frac{1}{2} \times} 2 \times | A E | = \textcolor{red}{\frac{1}{2} \times} 8$
$\implies | A E | = 4$

$\textcolor{w h i t e}{\times} \cos m \left(\angle D A E\right) = \frac{8}{4}$
$\textcolor{w h i t e}{\times \times \times \times \times x} = \frac{1}{2}$

$\implies \cos m \left(\angle D A E\right) = \cos 60$
$\implies \arccos \cos m \left(\angle D A E\right) = \arccos \cos 60$
$\implies m \left(\angle D A E\right) = 60$