Law of Sines/Cosines: A hill is inclined 5 degrees to the

Tigertigre2000

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A hill is inclined 5 degrees to the horizontal. A 45 ft pole stands at the top of the hill. How long a rope will it take to reach from the top of the pole to a point 35 ft. downhill from the base of the pole. (Insturctions say you either use law of sines, or law of cosines).

I don't know how to begin, or how to draw the picture.
All I know is that one angel is 5 degrees, I think one side is 35ft, and the pole is 45ft. but i don't know the other two angles, or the other two sides.

If you could help me out with this I'd really appriciate it.
 
Assuming I am interpreting the problem correctly:

Let DE be the distance from the base of the pole to our point E.

\(\displaystyle \L\\35sin(5)=3.05\)

\(\displaystyle \L\\35cos(5)=34.87\)

\(\displaystyle \L\\\sqrt{(48.05)^{2}+(34.87)^{2}}=59.37=CE\)

slopelr4.gif
 
Using Galactus' drawing, to use the law of cosines:

Let a be the length of the rope, b be the length of the pole (45) and c be the downhill slant length (35). Then A is the angle whos complement is angle DEC. Angle DEC will be 90-5=85 degrees. Thus Angle A will be 180-85=95 degrees.

You are looking for a.

So, \(\displaystyle a^2 = b^2 + c^2 - 2bc*Cos(A)\)
\(\displaystyle a^2 = 45^2 + 35^2 - 2(45)(35)Cos(95)\)
\(\displaystyle a^2 = 3250 - (-274.541) = 3524.541\)
\(\displaystyle a = 59.368\)
 
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