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Free 15th account... but there is a catch (easy)
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Free 15th account... but there is a catch (easy)Posted:
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Joined: Dec 06, 201013Year Member
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Status: Offline
Joined: Dec 06, 201013Year Member
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I will give you a free 15th prestige account if you can answer one of my old high school problems
I will give you 10 minutes and then if nobody has answerd I will put up the answer
Good Luck and Merry Christmas
a ship is coming into a harbor on an unusually high tide. The ship has to pass under the harbor bridge but the captain doesn't know if the ship will fit. he uses a theodolite to measure the angle at an unknown distance from the bridge and then re-measures the angle when he is 300 meters closer. The first angle measured is 2.3 degrees from sea level and the second angle is 3.3 degrees from sea level. If the ships height is 35metres out of the water, will it fit under the bridge. Show all working.
I will give you 10 minutes and then if nobody has answerd I will put up the answer
Good Luck and Merry Christmas
#2. Posted:
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Joined: Mar 13, 201014Year Member
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Joined: Mar 13, 201014Year Member
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Yes it will fit because the ship is MAGICAL
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#3. Posted:
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Joined: Dec 06, 201013Year Member
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Joined: Dec 06, 201013Year Member
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Haha. you were correct about it fitting but its not magical and you didnt show your work. the answer was
Height of bridge = y meters
Initial distance from bridge = x + 300 meters
Second distance from bridge = 300 meters
First measurement
tan(2.3) = y/(300 + x)
or 0.04 = y/(300 + x)
or 12 + 0.04x = y --(1)
Second measurement
tan(3.3) = y /x
0.06 = y/x
or y = 0.06x --(2)
Substitute the value of y from eq(2) in eq(1)
12 + 0.04x = 0.06x
or 0.02x = 12
or x = 600
Therefore y = 0.06*600 = 36 meters = height of the bridge
So height of the bridge is greater than height of the ship, which is 35 meters.
So the ship will fit.
Height of bridge = y meters
Initial distance from bridge = x + 300 meters
Second distance from bridge = 300 meters
First measurement
tan(2.3) = y/(300 + x)
or 0.04 = y/(300 + x)
or 12 + 0.04x = y --(1)
Second measurement
tan(3.3) = y /x
0.06 = y/x
or y = 0.06x --(2)
Substitute the value of y from eq(2) in eq(1)
12 + 0.04x = 0.06x
or 0.02x = 12
or x = 600
Therefore y = 0.06*600 = 36 meters = height of the bridge
So height of the bridge is greater than height of the ship, which is 35 meters.
So the ship will fit.
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#4. Posted:
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Joined: Dec 17, 200914Year Member
Posts: 27
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Status: Offline
Joined: Dec 17, 200914Year Member
Posts: 27
Reputation Power: 1
IdoFree15th wrote Haha. you were correct about it fitting but its not magical and you didnt show your work. the answer was
Height of bridge = y meters
Initial distance from bridge = x + 300 meters
Second distance from bridge = 300 meters
First measurement
tan(2.3) = y/(300 + x)
or 0.04 = y/(300 + x)
or 12 + 0.04x = y --(1)
Second measurement
tan(3.3) = y /x
0.06 = y/x
or y = 0.06x --(2)
Substitute the value of y from eq(2) in eq(1)
12 + 0.04x = 0.06x
or 0.02x = 12
or x = 600
Therefore y = 0.06*600 = 36 meters = height of the bridge
So height of the bridge is greater than height of the ship, which is 35 meters.
So the ship will fit.
yah and it can be found here all well [ Register or Signin to view external links. ]
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