the figure shows a 700 foot tower on the side of a hill that forms a 7° angle with the horizontal. find the length of each of the two guy wires that are anchored 50 feet uphill and downhill from the towers base and extend to the top of the tower.

Question

Accepted Answer

# Explanation:
## Step1: Determine the angles for the two triangles
For the uphill guy - wire:
The angle between the tower and the hill is $7^{\circ}$. The angle in the triangle formed by the tower, the uphill guy - wire, and the ground - like part is $90^{\circ}+7^{\circ}=97^{\circ}$
For the downhill guy - wire:
The angle in the triangle formed by the tower, the downhill guy - wire, and the ground - like part is $90^{\circ}-7^{\circ}=83^{\circ}$

## Step2: Use the Law of Cosines
The Law of Cosines is $c^{2}=a^{2}+b^{2}-2ab\cos C$
Let $a = 700$, $b = 50$
For the uphill guy - wire ($C = 97^{\circ}$):
$$
\begin{align*}
c_{1}^{2}&=700^{2}+50^{2}-2\times700\times50\times\cos(97^{\circ})\
c_{1}^{2}&=490000 + 2500-70000\times(-0.1219)\
c_{1}^{2}&=492500+8533\
c_{1}^{2}&=501033\
c_{1}&=\sqrt{501033}\approx708
\end{align*}
$$
For the downhill guy - wire ($C = 83^{\circ}$):
$$
\begin{align*}
c_{2}^{2}&=700^{2}+50^{2}-2\times700\times50\times\cos(83^{\circ})\
c_{2}^{2}&=490000+2500 - 70000\times0.1219\
c_{2}^{2}&=492500 - 8533\
c_{2}^{2}&=483967\
c_{2}&=\sqrt{483967}\approx696
\end{align*}
$$

# Answer:
The length of the uphill guy - wire is approximately $708$ feet and the length of the downhill guy - wire is approximately $696$ feet.

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QUESTION IMAGE

Question

Explanation:

Step1: Determine the angles for the two triangles

Step2: Use the Law of Cosines

Answer: