An airplane flies 200 km due west from city A to city B and then 260 km in the direction of 32.0° north of west from city B to city C.
(a) In straight-line distance, how far is city C from city A?
?km
(b) Relative to city A, in what direction is city C?
?° north of west
(a) Add up the x (east) and y (north) component displacement and then compute the magnitude of the resultant.
X = -200 - 260 cos 32
Y = 260 sin 32
distance = sqrt (X^2 + Y^2)
(b) The direction is in the second quadrant, since X is negative and Y is positive. If A is the angle north from west, A = arctan (-Y/X)
thank you for the help....but still could not figure out the answer
a) 484km
b) 18 degrees nw
(a) Well, it seems like city C is definitely taking the scenic route! To find the straight-line distance from city A to city C, we can use the good old Pythagorean theorem. So let's strap on our math helmets and solve this in no time!
The horizontal distance covered from A to B is 200 km, and the vertical distance covered from B to C is 260 km * sin(32.0°). To find the straight-line distance, we just need to combine these two distances using the Pythagorean theorem:
Distance AC = √((200 km)^2 + (260 km * sin(32.0°))^2)
Now, let's break out the calculators and find our answer!
(b) Now, for the direction of city C relative to city A. It is given that city C is 32.0° north of west from city B. So, to find the direction relative to city A, we just need to add 180° to that angle.
Direction of city C relative to city A = 32.0° + 180°
Alrighty, now we have all the components for our answer. Let's plug in the numbers and solve this puzzle!
To find the straight-line distance between city A and city C, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
(a) Let's break down the problem into two components: the horizontal (westward) distance traveled and the vertical (northward) distance traveled.
The horizontal distance from city A to city B is 200 km to the west.
The vertical distance traveled from city B to city C can be found by using the given information that the distance traveled in the direction 32.0° north of west is 260 km. To calculate the northward distance, we need to find the component of the distance traveled in the northward direction.
The northward component can be found by multiplying the total distance traveled (260 km) by the sine of the angle (32.0°). Using trigonometry, we can calculate this as:
Northward component = 260 km * sin(32.0°) = 136.36 km (rounded to two decimal places)
Now, we have the horizontal distance (200 km) and the northward distance (136.36 km). We can use the Pythagorean theorem to find the straight-line distance between city A and city C:
Straight-line distance = √((horizontal distance)^2 + (northward distance)^2)
Straight-line distance = √((200 km)^2 + (136.36 km)^2) ≈ 242.22 km (rounded to two decimal places)
Therefore, the distance between city A and city C is approximately 242.22 km.
(b) To find the direction of city C relative to city A, we need to determine the angle that the straight-line distance makes with the west direction. We can use trigonometry to find this angle.
The angle can be found by calculating the inverse tangent of the northward distance divided by the horizontal distance:
Angle = arctan(northward distance / horizontal distance)
Angle = arctan(136.36 km / 200 km) ≈ 35.97° (rounded to two decimal places)
Since city C is west of city A, the direction is measured as an angle north of the west direction.
Therefore, city C is approximately 35.97° north of west relative to city A.