9. a telephone pole is supported by two wires on opposite sides. at the top of the pole, the wires form an angle of 60°. on the ground, the ends of the wires are 15.0 m apart. one wire makes a 45° angle with the ground. how long are the wires, and how tall is the pole?

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Accepted Answer

# Explanation:
## Step1: Let the height of the pole be $ h $, the length of the wire making $ 45^{\circ} $ with the ground be $ x $, and the other wire length be $ y $.
From the right - triangle with the $ 45^{\circ} $ angle, we have $ \sin45^{\circ}=\frac{h}{x} $ and $ \cos45^{\circ}=\frac{\text{adjacent side}}{x} $. So, $ h = x\sin45^{\circ}=\frac{\sqrt{2}}{2}x $ and the adjacent side to the $ 45^{\circ} $ angle is $ x\cos45^{\circ}=\frac{\sqrt{2}}{2}x $.
Let the angle of the other wire with the ground be $ \theta $. Using the law of sines in the triangle formed by the two wires and the distance between their ground - ends ($ d = 15$ m). The sum of angles in a triangle is $ 180^{\circ} $, so if one angle is $ 60^{\circ} $ and we assume the angles at the ground are $ 45^{\circ} $ and $ \theta$, then $ \theta=180^{\circ}-(45^{\circ}+ 60^{\circ}) = 75^{\circ} $.
By the law of sines $ \frac{x}{\sin\theta}=\frac{y}{\sin45^{\circ}}=\frac{15}{\sin60^{\circ}} $
Since $ \sin60^{\circ}=\frac{\sqrt{3}}{2}$, $ \sin45^{\circ}=\frac{\sqrt{2}}{2}$, $ \sin75^{\circ}=\sin(45^{\circ}+30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4} $
From $ \frac{x}{\sin75^{\circ}}=\frac{15}{\sin60^{\circ}} $, we have $ x=\frac{15\sin75^{\circ}}{\sin60^{\circ}} $
$$
\begin{align*}
x&=\frac{15\times\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{3}}{2}}\
&=\frac{15(\sqrt{6}+\sqrt{2})}{4}\times\frac{2}{\sqrt{3}}\
&=\frac{15(\sqrt{2}+\ 1)\sqrt{3}}{2\sqrt{3}}\
&=\frac{15(\sqrt{2}+1)}{2}\approx15.5\space m
\end{align*}
$$
From $ \frac{y}{\sin45^{\circ}}=\frac{15}{\sin60^{\circ}} $, we have $ y = \frac{15\sin45^{\circ}}{\sin60^{\circ}} $
$$
\begin{align*}
y&=\frac{15\times\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}\
&=\frac{15\sqrt{2}}{\sqrt{3}}\
&=5\sqrt{6}\approx12.2\space m
\end{align*}
$$
## Step2: Calculate the height of the pole.
Since $ h = x\sin45^{\circ} $, substituting $ x=\frac{15(\sqrt{2}+1)}{2} $
$$
\begin{align*}
h&=\frac{15(\sqrt{2}+1)}{2}\times\frac{\sqrt{2}}{2}\
&=\frac{15(2 + \sqrt{2})}{4}\approx11.0\space m
\end{align*}
$$

# Answer:
The lengths of the wires are approximately $ 15.5\space m $ and $ 12.2\space m $, and the height of the pole is approximately $ 11.0\space m $

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Question

Explanation:

Step1: Let the height of the pole be \( h \), the length of the wire making \( 45^{\circ} \) with the ground be \( x \), and the other wire length be \( y \).

Step2: Calculate the height of the pole.

Answer: