Question 1078633: Hi, I need help on this question it is grade 11 math.
"A ship is tied to a dock with a rope of length 10 m. At low tide, the rope is stretched tight, forming an angle of 45 degrees with the horizontal. At high tide, the stretched rope makes an angle of 30 degrees with the horizontal . How much closer to the dock is, horizontally, is the ship at low tide than high tide? Determine an exact expression. Then use a calculator to determine an approximate answer, correct to the nearest tenth of a metre. "
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Helps to draw this.
cosine 45=x/10, where the hypotenuse is 10m of line, and the horizontal distance is the adjacent side
x=10cos 45=10*sqrt(2)/2=5 sqrt (2)=7.070 m
at high tide
x=10 cos 30=10*sqrt(3)/2=5 sqrt(3)=8.660 m
The difference is 8.66-7.07=1.59 m or 1.6 m.
The exact amount is 5(sqrt(3)-sqrt(2)) m
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