A plane flies in a direction of N70°E for 80 km and then on a bearing of S10°W for
150 km.
a How far is the plane from its starting point?
b What direction is the plane from its starting point?
I will convert the directions to vectors using conventional angles
first leg: (80cos20° , 80sin20°)
2nd leg: (150cos260° , 150sin260°)
resultant
= (80cos20° , 80sin20°) + (150cos260° , 150sin260°)
= (49.128, -120.3596)
magnitude = √(49.128^2 + (-120.3596)^2) = 130 km
direction:
tanØ = -120.3596/49.128
Ø = -67.8° or 292.2°
or
S 22.2° E
or
Make a sketch , let the original position be A, end of first leg B, and C the final position.
I have a triangle ABC with angle B = 60°
AB = 80 and BC = 150
by the cosine law:
AC^2 = 80^2 + 150^2 - 2(80)(150)cos60
= 16900
AC = √16900 = 130
by sine law:
sinA/150 = sin60/130
sinA = 150sin60/130 = .99926
angle A = 87.8°
looking at your diagram, subtracting 20° yields the same angle of 67.8° as in my first solution.
a) Oh, the poor plane! It's playing hide and seek with its starting point. Let's calculate its distance from the starting point using a bit of trigonometry. The first leg of the journey takes it N70°E for 80 km. This forms a right-angled triangle with the starting point as the origin, and the plane's final position as the hypotenuse. So, using some math magic, we can say that the horizontal distance covered by the plane is 80 km * cos(70°) and the vertical distance covered is 80 km * sin(70°). When we add these distances together, we get the total distance from the starting point.
b) Now, let's find out the direction the plane is from its starting point. We'll use a compass bearing for this one. The first leg of the journey takes it N70°E, which means it's 20° shy of due east. The second leg of the journey takes it on a bearing of S10°W, which means it's 20° shy of due south. Now, we need to do a bit of math to find the overall direction. Subtract 20° (shy of east) from 180° (shy of south) and you'll get the angle between the starting point and the final position. Oh, don't worry, I'll handle the math monster for you. Just give me a second.
Calculating... crunch, crunch, crunch...
The plane is approximately 100 km from its starting point, and its direction is about S60°W.
To find the distance traveled by the plane and its direction from the starting point, we can use vector addition.
a) To find the distance, we can use the Pythagorean theorem, as the traveled distance will form a right triangle.
The distance traveled towards N70°E for 80 km can be represented as a vector:
Vector1 = 80 km at N70°E
The distance traveled on a bearing of S10°W for 150 km can be represented as another vector:
Vector2 = 150 km at S10°W
To calculate the total distance traveled, we need to find the resultant vector of Vector1 and Vector2.
Using basic trigonometry, we can determine the horizontal and vertical components of each vector:
Vector1 (N70°E) has a horizontal component (x) and a vertical component (y):
x = 80 km * sin(70°)
y = 80 km * cos(70°)
Vector2 (S10°W) has a horizontal component (-x') and a vertical component (-y'):
x' = 150 km * sin(10°)
y' = 150 km * cos(10°)
Now, let's calculate the total horizontal and vertical components by adding the respective components of each vector:
Total horizontal component, x_total = x + (-x')
Total vertical component, y_total = y + (-y')
To find the total distance, D_total, we can use the Pythagorean theorem:
D_total = sqrt((x_total)^2 + (y_total)^2)
b) To find the direction, we can use trigonometry again to calculate the angle between the resultant vector and the positive x-axis:
Direction = tan^(-1)(y_total / x_total)
Now, let's calculate the distance and direction using the above formulas:
First, calculate the components:
x = 80 km * sin(70°) ≈ 72.62 km
y = 80 km * cos(70°) ≈ 29.85 km
x' = 150 km * sin(10°) ≈ 25.98 km
y' = 150 km * cos(10°) ≈ 148.07 km
Total horizontal component:
x_total = x + (-x') ≈ 72.62 - 25.98 ≈ 46.64 km
Total vertical component:
y_total = y + (-y') ≈ 29.85 - 148.07 ≈ -118.22 km
Now, calculate the total distance:
D_total = sqrt((x_total)^2 + (y_total)^2) ≈ sqrt((46.64)^2 + (-118.22)^2) ≈ sqrt(2172.77 + 13956.9) ≈ sqrt(16129.67) ≈ 127 km
Finally, calculate the direction:
Direction = tan^(-1)(y_total / x_total) ≈ tan^(-1)(-118.22 / 46.64) ≈ -69.13°
Thus, the answers are:
a) The plane is approximately 127 km from its starting point.
b) The plane is approximately N69.13°W from its starting point.
To find the distance and direction of the plane from its starting point, we can use vector addition.
Let's break down the information given:
1. The plane flew in a direction of N70°E for 80 km. This means the plane moved 80 km in a direction 70° east of north.
2. The plane then flew on a bearing of S10°W for 150 km. This means the plane moved 150 km in a direction 10° west of south.
Now, let's calculate the displacement vector for each leg of the plane's journey:
1. The North component of the first leg is 80 km * sin(70°) = 72.04 km North.
The East component of the first leg is 80 km * cos(70°) = 29.68 km East.
2. The South component of the second leg is 150 km * sin(10°) = 25.86 km South.
The West component of the second leg is 150 km * cos(10°) = 145.65 km West.
Next, we add the displacement vectors to get the net displacement of the plane:
North displacement = North component of first leg - South component of second leg
= 72.04 km - 25.86 km
= 46.18 km North
East displacement = East component of first leg - West component of second leg
= 29.68 km - (-145.65 km)
= 175.33 km East
Finally, we can find the distance and direction from the starting point using the Pythagorean theorem and trigonometry:
a) Distance = √(North displacement)^2 + (East displacement)^2
= √(46.18 km)^2 + (175.33 km)^2
≈ 184.9 km
b) Direction = atan2(East displacement, North displacement)
= atan2(175.33 km, 46.18 km)
≈ 75.69°
Therefore, the plane is approximately 184.9 km away from its starting point and is at a direction of about 75.69° from its starting point.