Chapter 6.2, Problem 49E

To determine

The length of the each of the two guy wires which are anchored 75 feet and downhill from the base of the tower.

Answer to Problem 49E

The length of the each two guys is 130.86ft and 118.84ft.

Explanation of Solution

Given:

The figure is given as:

  

Redrawing the given figure:

  

According to the law of the cosines, the triangle ABC with the sides is represented as:

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=a^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}$

Now, considering the triangle ADC, the law of the cosines is applied as:

  $\begin{array}{l}{d}^2=a^2+c^2-2ac\cos D\\ ={100}^2+{75}^2-2\left(100\right)\left(75\right)\cos {96}^0\\ =10000+5625-\left(15000\right)\left(-0.10\right)\\ =14125\end{array}$

Therefore, the side dis given as:

  $\begin{array}{l}d=\sqrt{17125}\\ =130.86\end{array}$

Now, considering the triangle BDC:

Applying the law of the cosines by substituting b=100, D=84-degre and c=75:

  $\begin{array}{l}{d}^2=b^2+c^2-2bc\cos D\\ ={100}^2+{75}^2-2\left(100\right)\left(75\right)\\ =10000+5626-15000\\ =14125\end{array}$

Hence, the side d is:

  $\begin{array}{l}d=\sqrt{14125}\\ =118.84\end{array}$

Thus, the length of the each two guys is 130.86ft and 118.84ft.