Trigonometry(3)quick 20 points

VINZ

2009-11-05 1:42 am

1.A countour map with scale of 1: 100000. The differences in height between every two consecutive contour lines are the same. the height of each contour line is not indicated, but a person knows from a sign of the road that the gradient of AB is 1 in 10. (AB= 1 and BC= 1.5 are measures on the map before.) Find the angle of inclination of BC.( coorect to 1 decimal place.)

2.Steven used the following method to find the height of the tower CD. He firstly moved to point A in front of the tower where the angle of elevation measured 45(degree) from point A to the top of the tower. He then moved 50m backward to point B where the angle of elevation to the top of the tower became 30 (degree). It is given that BAD is a straight line on the horizontal level. Find the height of the tower.

3. A submarine sails near an unmoving warship. It is given that submarine is 300m down the sea-level ans sails at a constant speed. If the angle of depression from the warship to the submarine is 8.5(degree) nowand 16.7(degree) two minutes later, find the speed of the submarine. (nearest 0.1 km/h)

4.From the deck of a boat, the angle of elevation of a helicopter is 53(degree). From the top of a mast which is 10m above the deck, the angle of elevation of the helicopter is 44(degree). How high is the helicopter from the deck?

5. From half-way up a lighthouse, the angle of depression of a boat is 23(degree). What wil the angle of depression be of the same boat from the top of the lighthouse?

回答 (1)

用戶9112

2009-11-05 3:25 am

✔ 最佳答案

1. I assume the point A, B and C are on consecutive contour lines.
Let the difference in height between every two consecutive contour lines be x
The horizontal distance between AB = 100,000
Gradient of AB is 1 in 10 => x/100,000 = 0.1 => x = 10,000
The horizontal distance between BC = 1.5 * 100,000 = 150,000
Angle of inclination is arctan x/150,000 = arctan 10,000/150,000 = 3.8 degrees
2. Let the height of the tower be x
Let AD be y
x/y = tan45 = 1 => x = y
x/(y + 50) = tan30 => x = (x + 50)(0.5774)
0.4226x = 28.87
x = 68.31m
The height of the tower is 68.31m
3. When the angle of depression = 8.5, the horizontal distance
= 300 / tan8.5
= 2007 m
When the angle of depression = 16.7, the horizontal distance
= 300 / tan16.7
= 1000 m
The submarine moved (2007 - 1000) = 1007 m in 2 minutes
Therefore speed of the submarine
= 1007m / 2min
= 1007 * 30 m/hr
= 30.2 km/h
4. Let the helicopter be x m above the deck
Let the horizontal distance between the deck and the helicopter be y m
x/y = tan53 => y = 0.7536x
(x - 10)/y = tan44
x - 10 = y tan44 = (0.7536x)(0.9657) = 0.7277x
0.2723x = 10
x = 36.72
The helicopter is 36.72 m above the deck
5. Let the height of the lighthouse be x m
Let the horizontal distance between the lighthouse and the boat be y m
0.5x/y = tan23 = 0.4245
x = 0.8489y
The angle of depression form the top of the lighthouse = arctan x/y
= arctan 0.8489y/y
= arctan 0.8489
= 40.33 degrees

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