A plane flies 400km North and then a further 600km or a bearing of 70 degree.
a) How far is the plane from the starting point.
b) On what bearing must the plane fly so that it returns to its original point.
c) How far north and how far east is the plane from the starting point.
Bearings are always measured clockwise from north.
A = start point
B = point when a plane was flying 400 km and starts to fly under a bearing of 70°
C = point after 600 km of fly
Angle betwen AB and BC = θ
A bearing of 70° mean θ = 180° - 70° = 110°
In this case you have a triangle:
AB = 400 km , BC = 600 km , θ = 110°
a)
Use law of cosines:
AC = AB² + BC² - 2 ∙ AB ∙ BC ∙ cos θ
AC² = 400² + 600² - 2 ∙ 400 ∙ 600 ∙ cos 110°
AC² = 160000 + 360000 - 480000 ∙ ( - 0.3420201433 )
AC² = 520000 + 480000 ∙ 0.3420201433
AC² = 520000 +164169.6688
AC² = 684169.6688
AC = √684169.6688
AC = 827.1454943 km
b)
The plane must bearing by angle α ( angle in point A )
Now law of sines:
sin α / sin θ = BC / AC
sin α / sin 110° = 600 / AC
sin α / 0.939692621 = 600 / 827.1454943
cross multiply
827.1454943 ∙ sin α = 600 ∙ 0.939692621
827.1454943 ∙ sin α = 563.8155726
sin α = 563.8155726 / 827.1454943
sin α = 0.681640142
α = arc sin ( 0.681640142 ) = 42.97194285°= 42° 58´ 19"
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Remark:
arc sin = sin⁻¹
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c)
In start point the plane it's not far from the north because it's flying to north.
E = far from the east
In point B angle between horizontal line and line BC is 90° - 70° = 20°
cos 20° = E / BC
far from the east:
E = BC ∙ cos 20° = 600 ∙ cos 20° = 600 ∙ 0.9396926208 = 563.81557248 km
a) Well, if the plane flies 400km north and then an additional 600km at a bearing of 70 degrees, we can calculate the total distance using some fancy math. Using the cosine rule, we can find that the total distance is approximately 891.84 kilometers. So, the plane is about 891.84 kilometers away from the starting point.
b) Ah, for the plane to steer its way back to the original point, it needs a bearing that creates a triangle with the starting point and the current position. To find this bearing, we subtract 70 degrees from 180 degrees (to account for the sharp turn) and then add another 180 degrees (because we're traversing the opposite direction). Therefore, the plane must fly at a bearing of 290 degrees to return to its original point.
c) To figure out how far north and how far east the plane is from the starting point, we need to use some trigonometry magic. Using sine and cosine, we can determine that the plane is approximately 688.88 kilometers north and 368.68 kilometers east from the starting point.
Remember, these calculations assume a perfectly clear sky with no distractions or detours from clowns like me!
I don't understand it
I don't understand it
To solve this problem, we can use basic trigonometry and vector addition. Let's break it down step by step:
a) How far is the plane from the starting point?
To find the distance from the starting point, we can use the Pythagorean theorem. The plane has traveled 400 km north, so we have a vertical distance of 400 km. The plane then travels a further 600 km at a bearing of 70 degrees.
To find the horizontal distance traveled, we can use trigonometry. We can decompose the 600 km into its east and north components. The east component can be found by multiplying 600 km by the cosine of the bearing (70 degrees). The north component can be found by multiplying 600 km by the sine of the bearing (70 degrees).
East component = 600 km * cos(70°)
North component = 600 km * sin(70°)
Now, we can find the total distance from the starting point using the Pythagorean theorem:
Distance = √(vertical distance^2 + horizontal distance^2)
= √(400 km^2 + (600 km * cos(70°))^2 + (600 km * sin(70°))^2)
Using a calculator, we can evaluate this expression to find the answer.
b) On what bearing must the plane fly so that it returns to its original point?
To determine the bearing to return to the original point, we need to consider the overall displacement of the plane. The vertical displacement is 400 km north, and the horizontal displacement is the east component minus the original 600 km.
Bearing = arctan(vertical displacement / horizontal displacement)
Using the values we have, we can calculate the bearing using a calculator.
c) How far north and how far east is the plane from the starting point?
To find how far north and east the plane is from the starting point, we can simply look at the vertical and horizontal components we calculated earlier:
North distance = 400 km + (600 km * sin(70°))
East distance = 600 km * cos(70°)
Using these formulas, we can find the specific distances.
I apologize for the confusion. Let's go through the problem step by step.
a) To find how far the plane is from the starting point, we can use the Pythagorean theorem. The plane flew 400km North, so it is 400km away from the starting point in the North direction. Then, it flew a further 600km on a bearing of 70 degrees.
To visualize this, imagine a right angle triangle:
- The 400km length forms the vertical side of the triangle.
- The 600km length forms the horizontal side of the triangle.
- The plane's position is the endpoint of the hypotenuse.
To find the length of the hypotenuse (the distance from the starting point to the plane's position), we can use the formula: c^2 = a^2 + b^2, where c is the hypotenuse and a and b are the lengths of the other two sides.
c^2 = 400^2 + 600^2
c^2 = 160000 + 360000
c^2 = 520000
c = sqrt(520000)
c ≈ 721.11 km
Therefore, the plane is approximately 721.11 km away from the starting point.
b) To find the bearing the plane must fly to return to its original point, we need to find the angle θ between the line connecting the starting point to the final position and the North direction.
Since the plane flew 600km on a bearing of 70 degrees, the angle θ is the complementary angle to 70 degrees, which is 180 - 70 = 110 degrees.
To return to the original point, the plane needs to fly in the opposite direction, which is 180 degrees from the current bearing. So, the plane needs to fly on a bearing of 180 + 110 = 290 degrees.
Therefore, the plane must fly on a bearing of 290 degrees to return to its original point.
c) To find how far north and how far east the plane is from the starting point, we can use trigonometry.
The plane flew 400km North, so it is 400km north of the starting point.
To find the east distance, we can use the angle of 70 degrees that the plane flew. We can break down the 600km distance into its north and east components using the sine and cosine functions.
The east component (E) is given by:
E = 600km * cos(70 degrees)
E ≈ 197.99 km
Therefore, the plane is approximately 400km north and 197.99 km east from the starting point.