document.write( "Question 714990: 17 From the top of a vertical cliff 68 m high, an observer
\n" ); document.write( "notices a yacht at sea. The angle of depression to the
\n" ); document.write( "yacht is 47è. The yacht sails directly away from the cliff,
\n" ); document.write( "and after 10 minutes the angle of depression is 15è. How
\n" ); document.write( "fast does the yacht sail?
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Algebra.Com's Answer #439253 by KMST(5328)\"\" \"About 
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\n" ); document.write( "T = top of the cliff (we assume eye of observer is there, 68 meters above water surface)
\n" ); document.write( "B = bottom of the cliff (at water level)
\n" ); document.write( "X = initial position of the yacht
\n" ); document.write( "Y = position of the yacht 10 minutes later
\n" ); document.write( "TH = a horizontal line, parallel to BY.
\n" ); document.write( "angle HTX = \"47%5Eo\" --> Angle BTX = \"90%5Eo-47%5Eo=43%5Eo\"
\n" ); document.write( "angle HTY = \"15%5Eo\" --> Angle BTY = \"90%5Eo-15%5Eo=75%5Eo\"
\n" ); document.write( "\"BX=68m%2Atan%2843%5Eo%29=68m%2A0.93255=63.4m\"
\n" ); document.write( "\"BY=68m%2Atan%2875%5Eo%29=68m%2A3.732=253.8m\"
\n" ); document.write( "The yacht covered
\n" ); document.write( "\"253.8m-63.4m=190.4m\" in 10 minutes.
\n" ); document.write( "We could say that its speed in meters per minute is \"190.4%2F10=19\"
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