Chapter 8.4, Problem 9CE

To determine

The value of x in the figure

Answer to Problem 9CE

The value of x is $9\sqrt{2}$

Explanation of Solution

Given information:

The given figure is

  

Let the diagonal or common hypotenuse of both triangle named h.

  $\begin{array}{l}\sin \theta =\frac{perpendicular}{hypotenuse}\\ \sin 60°=\frac{9\sqrt{3}}{h}\\ \frac{\sqrt{3}}{2}=\frac{9\sqrt{3}}{h}\\ h=18\end{array}$

  $\begin{array}{l}\cos \theta =\frac{base}{hypotenuse}\\ \cos 45°=\frac{x}{h}\\ \frac{1}{\sqrt{2}}=\frac{x}{18}\\ x=\frac{18}{\sqrt{2}}\\ x=9\sqrt{2}\end{array}$

Therefore, the value of x is $9\sqrt{2}$