A weather ballon is on a 1200 metre length string. Its angle of inclination from the horizontal is 78deg, and its bearing from due north is 25deg. Assuming that the string is perfecly straight and does not sag at all, then find how far north and how far east of its tether point the ballon is.

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- anonymous

is it an scalene triangle? cuz i think u need to use the pytagoras theorem to solve this one

- anonymous

hello??

- anonymous

If we want to find how far north and how far east, we have to convert the 1200 meter length not on the horizontal to a length on the horizontal. I wish I could draw it, but cos(78deg)=c/1200.
c=1200cos(78deg)=249.494 meters, this is the length of the directed line segment, and it represents the hypotenuse of the new triangle. This directed line segment goes into the first quadrant because the bearing is greater than 0 degrees but less than 90 degrees. The 25 degree angle is formed between the 249.494 meter line segment and the y-axis. The distance north represents the length in the y-direction, so cos(25deg) = b/249.494 --> b = 249.494cos(25deg)=226.118 meters north. The distance east represents the length in the x-direction, so sin(25deg) = a/249.494 --> a = 249.494sin(25deg) = 105.441 meters east.

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- anonymous

Appreciate all responses. I solved in the same way as branlegr, and checked using Pythag. It was confusing at first for me to perceive the shadow as the new hypotenuse, bit tricky. Really great site, didnt know these study sites existed. :)

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