Chapter 8.3, Problem 50AYU

To determine

Find the length of two guy wires?

Answer to Problem 50AYU

  $\boxed{501.28ft}$

Explanation of Solution

Given information:

A radio tower 500 feet high is located on the side of a hill with an inclination to the horizontal of $5°$ . How long should two guy wires be if they are to connect to the top of the tower and be secured at two points 100 feet directly above and directly below the base of the tower?

Calculation:

Here,we have AC=500 ft, BC=100 ft, CE=100 ft and $\angle CBD=\angle EBF=5°$

In triangle $\Delta BCD$ ,we have

  $\sin 5°=\frac{CD}{BC}$

  $\begin{array}{l}CD=BC\times \sin 5°\\ CD=100\times \sin 5°\\ CD\approx 8.72ft\mathrm{......}\left(1\right)\end{array}$

And,

  $\cos 5°=\frac{BD}{\begin{array}{l}BC\\ \end{array}}$

  $\begin{array}{l}BD=BC\times \cos 5°\\ BD=100\times \cos 5°\\ BD\approx 99.62ft\mathrm{.....}\left(2\right)\\ \end{array}$

From the figure, we have

  $AD=AC+CD$

  $AD\approx 500+8.72$

  $AD\approx 508.72ft\mathrm{......}\left(3\right)$

In the right triangle $\Delta ABD$ applying Pythagorean Theorem, we get

  ${\left(AB\right)}^2={\left(BD\right)}^2+{\left(AD\right)}^2$

  $\approx {\left(99.62\right)}^2+{\left(508.72\right)}^2$ using (2) and (3)

  $\approx 268720.1828$

  $AB\approx 518.38ft\mathrm{.....}\left(4\right)$

Hence,length of the one guy wire (AB) is approximately 518.38ft

Now in triangle $\Delta BCD$ , we have

  $\angle CBD+\angle BDC+\angle DCB=180°$

  $\angle DCB=180°-\left(\angle CBD+\angle BDC\right)$

  $\angle DCB=180°-\left(5°+90°\right)$

  $\angle DCB=85°\mathrm{.....}\left(5\right)$

Thus, $\angle ACE=\angle DCB=85°$

Now in the triangle $\Delta ACE$ , we have $AC=500ft$ , $CE=100ft$ and $\angle ACE=50°$

Here, two sides and one angle is given, it is an SSA problem in which given angle is not apposite to the given sides. By using the Law of Cosines, we get

  ${\left(AE\right)}^2={\left(AC\right)}^2+{\left(CE\right)}^2-2\left(AC\right)\left(CE\right)\cos \left(\angle ACE\right)$

  $={\left(500\right)}^2+{\left(100\right)}^2-2\left(500\right)\left(100\right)\times \cos 85°$

  $=250000+10000-100000\times \cos 85°$

  $\approx 251284.4257$

  $AE\approx 501.28ft$

Hence, length of the other guy wire (AB) is approximately $\boxed{501.28ft}$