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I am having trouble drawing the picture that goes with this problem:

A $100 ft$ vertical tower is to be erected on the side of a hill that makes a $6$ degree angle with the horizontal. Find the length of the guy wire that will be anchored $75 ft$ downhill from the base of the tower.

Can anyone describe how the picture will look like?

Thanks

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3 Answers 3

up vote 4 down vote accepted

I think picture can look like this:

enter image description here

$OS$ $-$ tower, $BS$ (red line) $-$ guy wire, anchored downhill.

$OS$ = 100 ft, $OB$ = 75 ft, $\angle SOB = 96^\circ$.

You can use Law of cosines to find this guy wire length.

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If the ground were level, the guy-wires on either side of the tower would have the same length. Instead, the ground is sloping at a 6º angle from level. A wire from the top of the tower to the side of the slope "uphill" from the tower will be shorter than the wire from the tower top to the ground "downhill" from the tower. The "downhill" wire is attached to the ground at a measured distance along the ground of 75 feet. If the tower itself is 100 feet tall, how long does the wire on the downhill side need to be in order to reach the ground (when pulled taut)?

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I cannot draw a picture, so will use a thousand words instead.

Draw a horizontal line (the $x$-axis, say). Now draw a line, say through the origin, sloping fairly gently upwards. An angle of $6^\circ$ may not sound like much, about a $1$ foot rise every $10$ feet. But it is not much fun at the end of a race.

Some distance along the line, draw a point $B$, the bottom of the tower. The tower is a vertical line. Let its top be $T$. The length of $BT$ is $100$.

Downhill from $B$ is one of the points where we anchor the towed. (We hope it is also anchored other places. Let this anchor point be $A$. We are told that $AB=75$.

Finally, join $A$ and $T$ by a straight line. You are asked to find the length of $AT$.

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