Question

1. two wires are supporting a tent pole, as shown.6 a) how far apart are the wires fixed in the ground, to the nearest tenth of a meter? b) find the angle each wire makes with the ground, to the nearest degree. state any assumptions you make.

Explanation:

Step1: Use the Law of Cosines for part a

Let \(AB = 2.2\) m, \(AC=2.8\) m and \(\angle BAC = 80^{\circ}\). The Law of Cosines states that \(BC^{2}=AB^{2}+AC^{2}-2\cdot AB\cdot AC\cdot\cos\angle BAC\).
Substitute the values: \(BC^{2}=2.2^{2}+2.8^{2}-2\times2.2\times2.8\times\cos80^{\circ}\)
\[

$$\begin{align*} BC^{2}&=4.84 + 7.84-12.32\times0.173648\\ BC^{2}&=4.84 + 7.84 - 2.13945\\ BC^{2}&=10.54055\\ BC&\approx3.2\text{ m} \end{align*}$$

\]

Step2: Use the Law of Sines for part b

Let's assume the tent - pole is perpendicular to the ground. Let the angle that the wire \(AB\) makes with the ground be \(\theta_{1}\) and the angle that the wire \(AC\) makes with the ground be \(\theta_{2}\).
First, find the area of \(\triangle ABC\) using \(S=\frac{1}{2}AB\cdot AC\cdot\sin\angle BAC\).
\[

$$\begin{align*} S&=\frac{1}{2}\times2.2\times2.8\times\sin80^{\circ}\\ S&=\frac{1}{2}\times2.2\times2.8\times0.984808\\ S&=3.03394 \end{align*}$$

\]
Also, \(S = \frac{1}{2}\times BC\times h\), where \(h\) is the height of the tent - pole. Since \(BC\approx3.2\) m, \(h=\frac{2S}{BC}=\frac{2\times3.03394}{3.2}\approx1.896\) m.
For the wire \(AB = 2.2\) m, \(\sin\theta_{1}=\frac{h}{AB}=\frac{1.896}{2.2}\approx0.8618\), so \(\theta_{1}\approx59^{\circ}\).
For the wire \(AC = 2.8\) m, \(\sin\theta_{2}=\frac{h}{AC}=\frac{1.896}{2.8}\approx0.6771\), so \(\theta_{2}\approx43^{\circ}\).

Answer:

a) \(3.2\) m
b) The wire of length \(2.2\) m makes an angle of approximately \(59^{\circ}\) with the ground and the wire of length \(2.8\) m makes an angle of approximately \(43^{\circ}\) with the ground. Assumption: The tent - pole is perpendicular to the ground.