The pilot of an airplane left A and flew 200 miles in the direction S 20deg20mins W. He then turned and flew 148 miles in the direction S 69deg40mins E. If he now heads back to A, in what direction should he fly?
The pilot travels 200 miles S at
20 deg.-20 min. When he changes direction, he makes a 90 deg angle
(20 deg.-20min.) + (69deg.-40 min.) =
90 Deg. During his return path, he completes the rt. triangle.
TanA = 200/148 = 1.3514, A = 53.5 deg.
B = 90 - 53.5 = 36.5 deg. = 36 Deg,30min.
Return Dir. = 36 Deg,30min - 20deg,20
=N 16deg.,10 min. W.
To find the direction in which the pilot should fly back to point A, we can use the concept of vector addition.
Let's break down the two legs of the journey:
First leg: The pilot flew 200 miles in the direction S 20°20' W.
Second leg: The pilot flew 148 miles in the direction S 69°40' E.
In order to determine the overall direction, we can add these two vectors together. To do so, we need to convert the given directions into components (north-south and east-west directions).
For the first leg:
- The south component is 200 * sin(20°20'W) ≈ -66.45 miles (negative sign indicates southward direction).
- The west component is 200 * cos(20°20'W) ≈ -188.60 miles (negative sign indicates westward direction).
For the second leg:
- The south component is 148 * sin(69°40'E) ≈ -70.99 miles (negative sign indicates southward direction).
- The east component is 148 * cos(69°40'E) ≈ 57.55 miles (positive sign indicates eastward direction).
Now, let's add up the components:
- North-South component: -66.45 miles + (-70.99 miles) ≈ -137.44 miles.
- East-West component: -188.60 miles + 57.55 miles ≈ -131.05 miles.
To find the total distance, we can use the Pythagorean theorem:
Total distance = √((-137.44 miles)^2 + (-131.05 miles)^2)
≈ √(18905.77 miles^2 + 17160.30 miles^2)
≈ √(360659237.35 + 294417630.25)
≈ √(655076867.60)
≈ 25564.02 miles.
Finally, to find the direction:
Direction = arctan(East-West component / North-South component)
= arctan(-131.05 miles / -137.44 miles)
≈ arctan(0.953)
≈ 43.11°.
Therefore, the pilot should fly in the direction approximately 43° (or N 43° E) to head back to point A.
To find the direction in which the pilot should fly back to point A, we can use vector addition.
Let's represent the initial displacement as vector A. The magnitude of vector A is 200 miles, and its direction is S 20°20' W.
Similarly, let's represent the second displacement as vector B. The magnitude of vector B is 148 miles, and its direction is S 69°40' E.
Now, to find the net displacement, we'll add the two vectors together.
To add vectors, we need to resolve them into their horizontal (East-West) and vertical (North-South) components.
For vector A, the horizontal component can be found using the formula:
Ax = A * sin(θ)
where A is the magnitude of vector A and θ is the angle between the vector and the East-West direction.
Similarly, the vertical component of vector A can be found using the formula:
Ay = A * cos(θ)
Using the same formulas for vector B, we can find its horizontal and vertical components, Bx and By, respectively.
Now, let's add the horizontal and vertical components of vectors A and B to get the net horizontal (Rx) and vertical (Ry) components of the displacement.
Rx = Ax + Bx
Ry = Ay + By
Next, we'll find the magnitude (R) and direction (θ) of the net displacement using the Pythagorean theorem and inverse trigonometric functions.
R = √(Rx^2 + Ry^2)
θ = tan^(-1)(Ry/Rx)
Finally, we'll convert the direction angle θ from the standard mathematical convention (measured from the positive x-axis counterclockwise) to the compass direction convention.
For example, if θ is positive, the direction will be measured counterclockwise from the East. If θ is negative, the direction will be measured clockwise from the East. Add or subtract θ from 90° to get the proper compass direction.
Note: It is important to convert degrees, minutes, and seconds into decimal form (degrees + minutes/60 + seconds/3600) before performing calculations.
By following these steps, you can find the direction in which the pilot should fly back to point A.