Very hard trigonometry question? - theodolite measure
a ship in a port of the exceptionally high tides. The boat has passed under the bridge port, do not know but the captain, is likely where the vessel. With a theodolite, the angle in a disputed stretch of the bridge and back to the angle that measures, measure 300 meters. The first measured angle is 2.3 degrees in sea level and the second angle of 3.3 degrees sea level. If the height is usually about 35metres of water, fits under the bridge. They all work.
Monday, January 11, 2010
Theodolite Measure Very Hard Trigonometry Question?
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Meters from the bridge, and =
ReplyDeleteGetting away from the bridge = x + 300 m
Second distance between the deck = 300 m
First measure
Tan (2,3) = Y / (300 + x)
or 0.04 = Y / (300 + x)
or 12 + y = 0.04x - (1)
Second measurement
Tan (3,3) = Y / X
0.06 = Y / X
y = 0.06x - (2)
Change the value and the equation (2) in equation (1)
12 + 0.04x = 0.06x
0.02xo = 12
ox = 600
So Y = 0.06 * 600 = 36 m = height of the bridge
So the bridge is greater than the height of the nave, the 35 meters.
Thus, fit the ship.
H is the height of the bridge.
ReplyDeleteH (cot2.3 - cot3.3) = 300
The solution for H
H = 300 / (cot2.3 - cot3.3) = 39.7 m> 35 m
Therefore, the boat must conform to the walkway.
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Ideas, to save time: Use cotangent.
H is the height of the bridge.
ReplyDeleteH (cot2.3 - cot3.3) = 300
The solution for H
H = 300 / (cot2.3 - cot3.3) = 39.7 m> 35 m
Therefore, the boat must conform to the walkway.
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Ideas, to save time: Use cotangent.
The bridge is 39,7 m above the point where action is taken. Thus, fit the ship.
ReplyDeleteDraw the triangle and you do not have points of view:
2.3, 1, 176.7
The opposite side of the angle of 1 degree is 300 meters long.
Use the sine rule to find other lengths HTE.
Then you will find the vertical component to 39.7 m. Works