Step 1

Given that:

A pole has two wires attached to it, one on side forming two right triangles as shown,

From the figure,

Length of pole = OB

\(\displaystyle\angle{O}{A}{B}={41}^{{\circ}}\)

\(\displaystyle\angle{O}{B}{A}={90}^{{\circ}}\)

\(\displaystyle{A}{B}={34}{f}{t}\)

length of wire 1= OA

Step2

Part A

To calculate the length of pole,

In \(\triangle OAB\),

\(\displaystyle{\frac{{{P}{e}{r}{p}{e}{n}{d}{i}{c}ul{{a}}{r}}}{{{B}{a}{s}{e}}}}={\tan{{\left({41}^{{\circ}}\right)}}}\)

\(\displaystyle{\frac{{{O}{B}}}{{{A}{B}}}}={\tan{{\left({41}^{{\circ}}\right)}}}\)

\(\displaystyle{O}{B}={A}{B}\times{\tan{{\left({41}^{{\circ}}\right)}}}\)

\(\displaystyle={34}\times{\tan{{\left({41}^{{\circ}}\right)}}}\)

\(\displaystyle={29.56}{f}{t}\)

Hence the poll is 29.56 ft tall

Step3

Part B

To calculate the length of wire 1,

Apply Pythagoras theorem in triangle OAB,

In \(\triangle OAB\),

\(\displaystyle{\left({O}{A}\right)}^{{{2}}}={\left({O}{B}\right)}^{{{2}}}+{\left({A}{B}\right)}^{{{2}}}\)

\(\displaystyle={\left({29.56}\right)}^{{{2}}}+{\left({34}\right)}^{{{2}}}\)

\(\displaystyle={873.7936}+{1156}\)

\(\displaystyle{\left({O}{A}\right)}^{{{2}}}={2029.7936}\)

\(\displaystyle{O}{A}=\sqrt{{{2029.7936}}}\)

\(\displaystyle{O}{A}={45.05}{f}{t}\)

Hence wire 1 is 45.05 ft long