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1. A young tree is supported by two wires one on either side of the tree, that meet at the trees trunkand form a 80°angle. The longer wire connected to a stake on the right side of the tree and makesa 40° with the ground. If the stake holding the shorter wire is 12 feet away from the stake holdingthe longer wires in the ground, how long is each wire?

Sagot :

From the diagram provided in the question, we have the following triangle:

We have the following angle and side measures:

[tex]\begin{gathered} z=80\degree \\ y=40\degree \\ G=12\text{ feet } \\ x=180-80-40=60\degree\text{ (Sum of angles in a triangle)} \end{gathered}[/tex]

Recall the Sine Rule:

[tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}[/tex]

Comparing to our triangle, we have the sine rule applied as:

[tex]\frac{L}{\sin x}=\frac{G}{\sin z}=\frac{S}{\sin y}[/tex]

Length of Longer Wire (L):

[tex]\frac{L}{\sin x}=\frac{G}{\sin z}[/tex]

Substituting the given values, we have:

[tex]\begin{gathered} \frac{L}{\sin60}=\frac{12}{\sin 80} \\ L=\frac{12\sin 60}{\sin 80} \\ L=10.55\text{ feet} \end{gathered}[/tex]

The length of the longer wire is approximately 10.6 feet.

Length of Shorter Wire (S):

[tex]\frac{G}{\sin z}=\frac{S}{\sin y}[/tex]

Substituting known values, we have:

[tex]\begin{gathered} \frac{12}{\sin80}=\frac{S}{\sin 40} \\ S=\frac{12\sin 40}{\sin 80} \\ S=7.83\text{ feet} \end{gathered}[/tex]

The length of the shorter wire is approximately 7.8 feet.

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