An aeroplane is flying horizontally directly towards the city at an altitude of 400 metres. At a given time the pilot views the city lights of Melbourne at an angle of depression of 1.5 degrees. Two minutes later the angle of depression of the city lights is 5 degrees. Find the speed of the aeroplane in km/hr correct to one decimal.
A diagram and review of the basic trig functions will show that the distance d flown by the plane between observations is
d = 400 cot 1.5° - 400 cot 5° = 10700m = 10.7km
10.7km/2min * 60min/hr = 321 km/hr
To solve this problem, we can use trigonometry and the concept of similar triangles.
Let's assume the speed of the airplane is "v" km/hr.
From the given information, we have two triangles: the first triangle when the angle of depression is 1.5 degrees and the second triangle when the angle of depression is 5 degrees.
Let's label the important points:
- A: the altitude of the airplane (400 meters)
- B: the position of the observer
- C: the city lights (ground level)
- D: the position of the observer 2 minutes later (same altitude as B)
- E: the position of the observer 2 minutes later (parallel to D)
- F: the point below C vertically aligned with D and E
In triangle ABC, we have:
tan(1.5 degrees) = AB / BC
In triangle ADE, we have:
tan(5 degrees) = AD / DE
Both triangles share side AC, so we can equate the two equations:
AB / BC = AD / DE
Using the fact that the distance covered by the airplane in 2 minutes is equal to the ground distance covered by the observer, we have:
AB = v * 2/60 (since 2 minutes is 2/60 hours)
We can substitute AB in terms of v:
(v * 2/60) / BC = AD / DE
Since DE = BC (as they are parallel), we can simplify the equation:
v * 2/60 = AD / BC
Now, let's plug in the values:
400 / BC = tan(5 degrees) / tan(1.5 degrees)
Solving for BC, we get:
BC = 400 / (tan(5 degrees) / tan(1.5 degrees))
BC ≈ 6260.274 meters
To convert meters to kilometers, we divide BC by 1000:
BC ≈ 6.260274 kilometers
Now, we can solve for v:
v * 2/60 = 400 / BC
v ≈ (400 / BC) * (60/2)
v ≈ 24.070606 km/hr
Therefore, the speed of the airplane is approximately 24.1 km/hr.
To find the speed of the airplane, we need to use trigonometry and the concept of angular velocity. Here's how we can solve the problem step by step:
Step 1: Draw a diagram:
Start by drawing a right-angled triangle to represent the situation. Label the altitude of the airplane as 400 meters, the angle of depression at the first observation as 1.5 degrees, and the angle of depression at the second observation as 5 degrees.
C
|\
| \
| \
| \ 400 meters
| \
| \
B------A (city lights of Melbourne)
Step 2: Identify the relevant trigonometric ratios:
We can use the tangent function to solve this problem, as the tangent of an angle is defined as the opposite side divided by the adjacent side of a right triangle.
tan(angle) = opposite / adjacent
In this case, the opposite side is the altitude of the airplane (400 meters) and the adjacent side is the horizontal distance traveled by the airplane.
Step 3: Calculate the horizontal distance at the first observation:
Let's denote the first observation as point X on the ground. From point X to point A (the city lights), the airplane has covered a certain horizontal distance in two minutes. We can call this distance "x" (in meters).
From the trigonometric relation mentioned above, we can write:
tan(1.5 degrees) = 400 / x
Using a scientific calculator, find the value of tan(1.5 degrees), which is approximately 0.0262. Rearrange the equation to solve for x:
x = 400 / tan(1.5 degrees)
≈ 400 / 0.0262
≈ 15267.2 meters
Step 4: Calculate the horizontal distance at the second observation:
Similarly, at the second observation, the airplane has covered a horizontal distance "y" (in meters) from point X to point A (the city lights) in two minutes. We can use the trigonometric relation and the given angle to find the value of y:
tan(5 degrees) = 400 / y
Using a scientific calculator, find the value of tan(5 degrees), which is approximately 0.0875. Rearrange the equation to solve for y:
y = 400 / tan(5 degrees)
≈ 400 / 0.0875
≈ 4571.4 meters
Step 5: Calculate the change in distance:
The change in distance between the two observations is given by the difference in the horizontal distances covered by the airplane, which can be calculated as:
change in distance = y - x
≈ 4571.4 meters - 15267.2 meters
≈ -10695.8 meters (Note: Negative sign indicates the airplane is getting closer to the city)
Step 6: Calculate the speed of the airplane:
The speed of the airplane can be calculated by dividing the change in distance by the time taken. In this case, the time taken is two minutes, which is equivalent to 1/30 hours.
speed = change in distance / time
= (-10695.8 meters) / (1/30 hours)
= -320874 meters/hour
Step 7: Convert the speed to km/hr:
To convert the speed from meters/hour to km/hr, divide the speed by 1000:
speed = -320874 meters/hour = -320.874 km/hr
Step 8: Round the answer to one decimal place:
Since the problem asks for the answer to be correct to one decimal place, round the speed to one decimal place:
speed ≈ -320.9 km/hr
Therefore, the speed of the airplane is approximately 320.9 km/hr (rounded to one decimal place). Note that the negative sign indicates the airplane is flying towards the city.
I drew a horizontal flight path 400 m above the ground.
On that path, I labeled the first position of the plane as A and the 2nd position as B, and the location of Melbourne on the ground as M
I also labeled C on the flight path as the point directly above M.
Let the speed of the plane be v km/h
then AB = (1/30)v or v/30
In triangle ABM, angle AMB = 3.5°
by the sine law:
BM/sin 1.5° = (v/30) /sin 3.5°
BM = (v/30)sin 1.5° / sin 3.5
in the right-angled triangle BCM,
sin 5° = .4/BM , (400 m = .4 km)
BM = .4/sin 5
(v/30)sin 1.5° / sin 3.5 = .4/sin 5
v/30 = (.4/sin5)(sin3.5/sin1.5)
v = 30(.4/sin5)(sin3.5/sin1.5)
= .....