SOLUTION: there is to be a water fountion in the back yard of the buliding. the water tap is 25 feet away from the proposed fountaion location at a bearing of 75 dregress. however becaue of

Algebra ->  Trigonometry-basics -> SOLUTION: there is to be a water fountion in the back yard of the buliding. the water tap is 25 feet away from the proposed fountaion location at a bearing of 75 dregress. however becaue of       Log On


   

Question 810532: there is to be a water fountion in the back yard of the buliding. the water tap is 25 feet away from the proposed fountaion location at a bearing of 75 dregress. however becaue of the construction of the garage the water pipe extends 10 feet out at a 90 dergree bearing before turning towards the fountaion . what is the length of the pipe needed from the end of the fountain?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
there is to be a water fountain in the back yard of the buliding. the water tap is 25 feet away from the proposed fountain location at a bearing of 75 dregress. however because of the construction of the garage the water pipe extends 10 feet out at a 90 degree bearing before turning towards the fountain . what is the length of the pipe needed from the end of the fountain?
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Draw the picture on a coordinate system.
Put the tap (T) at the origin.
Mark an angle of 75degrees up from the positive x-axis.
Let the fountain(F) be on that 75 degree bearing line and 25 ft. from T.
Draw a line segment from T to (0,10) on the y-axis.
Draw a line segment from (0,10) to F and label it "x".
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Solve for "x" using the Law of Cosines::
x^2 = 10^2+25^2-2*10*25*cos(15 degrees)
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x^2 = 235.93
x = 15.36 ft
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Total pipe needed: 10'+15.36 = 25.36 ft.
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Cheers,
Stan H.
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