Chapter 9.10, Problem 5PSA

To determine

To find:The height of the trapezoid ABCD .

Answer to Problem 5PSA

The height of the trapezoid ABCD is $DE=15.4$

Explanation of Solution

Given information:The following information has been given

  $AD=CB=25$

  $\angle DAE={38}^0$

Formula used:Basic sine identity has been used

Calculation:In the right triangle ADE, we can say that

  $\sin {38}^0=\frac{DE}{AD}$

Now, from the table of trigonometric values, we can say that

  $\sin {38}^0=0.6157$

Now, comparing the two values, we get

  $0.6157=\frac{DE}{AD}=\frac{DE}{25}\to DE=15.4$

Hence, we can say that

  $DE=15.4$