Chapter 9.9, Problem 15PSB

To determine

To find:The sine of one of the acute angles in the trapezoid

Answer to Problem 15PSB

The value of $\sin \angle \theta =\frac{4}{5}$

Explanation of Solution

Given information:We know by the given values that the parallel sides are of measure five and seventeen each.

Formula used:Basic trigonometric identity, and Pythagoras theorem has been used

Calculation:For finding the sine of the acute angle, we draw an altitude to the trapezoid from the vertex of the smaller parallel side to the larger one.

Now, we know that the remaining side of the triangle will be

  $\frac{1}{2}(17-5)=6$

And, the hypotenuse is given as ten units.

Now, we find the remaining side by applying Pythagoras in the right triangle, we get

  $6^2+x^2={10}^2$

Thus, we get

  $x=8$

Now, we know that value of sine angle will be

  $\sin \angle \theta =\frac{8}{10}=\frac{4}{5}$