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4 ftA wire isneeded tosupport averticalpole 4 feettall. Thecable willbeanchoredto a stake3 feet fromthe baseof thepole. How3 ftA cable of lengthfeet is needed.

4 ftA wire isneeded tosupport averticalpole 4 feettall. Thecable willbeanchoredto-example-1
User Sachin J
by
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1 Answer

1 vote

SOLUTION

From the image in the question, we can see that it is a right triangle problem.

And we can use the Pythagoras theorem to solve for the length of cable wire needed.

Pythagoras theorem states that:


\begin{gathered} \text{hypotenuse}^2=opposite^2+adjacent^2 \\ x^2=4^2+3^2 \end{gathered}
\begin{gathered} x^2=16+9 \\ x^2=25 \\ \text{Take the square roots of both sides:} \\ \sqrt[]{x^2}=\sqrt[]{25} \\ x=5ft \end{gathered}

Final Answer:

A cable of length 5 feet is needed.

4 ftA wire isneeded tosupport averticalpole 4 feettall. Thecable willbeanchoredto-example-1
User Manuel R
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4.3k points