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 hight 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 hight tide? Determine an exact expression. Then use a calculator to determine an approximate aanswer, corect to the nearest tenth of a metre.
At low tide the rope/ship/dock system forms a 45-45-90 triangle with hypotenuse of 10. Then the sides of the triangle are `5sqrt(2)` each.
At low tide the boat is `5sqrt(2)` meters from the dock.
At high tide the rope/ship/dock system forms a 30-60-90 right triangle with hypotenuse 10; the shorter leg the height of the dock above the water, and the longer leg the horizontal distance to the boat. The longer leg is `5sqrt(3)` .
At high tide the boat is `5sqrt(3)` meters from the dock.
The boat is `5sqrt(3)-5sqrt(2)` meters closer to the dock at low tide.
`5sqrt(3)-5sqrt(2)~~1.589` so a calculator approximation is 1.6meters.
It should be noted that in the high tide the sea level is high and it comes closer to the deck. So the rope angle that makes with the horizontal is high which is denoted by `anglePOQ` . The same at low tide is denoted by `angleAOB` .
`anglePOQ = 45`
`angleAOB = 30`
`OQ = OB = 10`
At low tide;
`(OA)/(OB) = cos30`
`OA = 10cos30`
At high tide;
`(OP)/(OQ) = cos45`
`OP = 10cos45`
If ship is xm closer to the deck in high tide than low tide then;
`x = OA-OP`
`x = 10(cos30-cos45)`
So the exact expression required is `10(cos30-cos45)`
`x = 10(cos30-cos45)`
`x = 1.589m`
The required answer to the nearest tenth of a meter is 1.6m.
Consider a right triangle whose hypotenuse is the rope, and the height and distance are the legs.
Using SOHCAHTOA, we know the following:
`cos(30^o) = y/10`
where x is the distance of the ship from the dock during low tides, and y the distance of the ship from the dock during high tides.
We want to get the diffence in the distance y - x (since it will be farther away during high tides).
Using the equations above:
`x = 10cos(45^o)`
`y = 10cos(30^o)`
`y - x = 10(cos(30^o) - cos(45^o))`
Hence, the exact value of the distance is `10(cos30^o - cos45^o)`
Using a calculator, this is approximately 1.59 meters.