Chapter 9.6, Problem 43PS

Solution

To find: the angle $\theta$ will the two trapeze artists meet.

The two trapeze artists will meet at angle of ${119.54}^{\circ }.$

Given:

The diagram is shown below:

  

Calculation:

From the figure it is see that the flyer on the left bar makes contact with the flyer on the right bar at point $C.$ This situation is possible only if the distance between $A$ and $D$ is same as the distance between $A$ and $C.$

Hence,

  $\begin{array}{l}a=AC\\ a=AD\\ a=17\end{array}$

Similarly, the distance between $B$ and $E$ must be same as the distance between $B$ and $C.$

Hence,

  $\begin{array}{l}b=BE\\ b=BC\\ b=13\end{array}$

In a triangle, the law of cosines says:

  $c^2=a^2+b^2-2ab\cos \theta$

Substitute the values of $a,b\text{ and }c$ as:

  $\begin{array}{c}{\left(26\right)}^2={\left(17\right)}^2+{\left(13\right)}^2-2\left(17\right)\left(13\right)\cos \theta \\ 676=289+169-442\cos \theta \\ 676-458=-442\cos \theta \\ -442\cos \theta =218\\ \cos \theta =-\frac{218}{442}\\ \cos \theta =-0.493\\ \theta ={\cos }^{-1}\left(-0.493\right)\\ \theta ={119.537}^{\circ }\\ \theta \approx {119.54}^{\circ }\end{array}$

Hence, the two trapeze artists will meet at angle of ${119.54}^{\circ }.$