A ship leaves a port and travels 21 km on a bearing of 032° and then 45 km on a bearing of 287°. (a) calculate it's distance from the port. (b) calculate the bearing of the port from the ship
https://www.jiskha.com/questions/1786587/a-ship-leaves-port-and-travels-21km-on-a-bearing-of-o32-degrees-and-then-45km-on-a-bearing
All angles are measured CW from +y-axis.
a. PB = PA+AB = 21km[32o]+45km[287o].
PB = (21*sin32+45*sin287)+(2 1*cos32+45*cos287)i
PB = -32 + 31i = 44.6km[-46o] = 44.6km[46o] W. of N. = 314o CW.
b. Bearing of BP = 314-180 = 134o.
To solve this problem, we can use the distance formula and the sine and cosine rules of triangles. Let's calculate the distance from the port and the bearing of the port from the ship.
(a) Distance from the port:
First, let's calculate the horizontal and vertical components of the ship's motion using trigonometry.
Horizontal Component:
Distance = 21 km
Bearing = 032°
To find the horizontal component, we use the cosine rule:
cos(θ) = adjacent/hypotenuse
cos(032°) = horizontal component/21 km
horizontal component = 21 km * cos(032°) ≈ 20.37 km
Vertical Component:
Distance = 45 km
Bearing = 287°
To find the vertical component, we use the sine rule:
sin(θ) = opposite/hypotenuse
sin(287°) = vertical component/45 km
vertical component = 45 km * sin(287°) ≈ -28.98 km
Note: The sine of 287° is negative as it is in the downward direction.
Now let's calculate the total displacement from the port using the horizontal and vertical components:
Total Displacement = √[(Horizontal Component)^2 + (Vertical Component)^2]
Total Displacement = √[(20.37 km)^2 + (-28.98 km)^2]
Total Displacement ≈ 35.55 km
Therefore, the ship is approximately 35.55 km from the port.
(b) Bearing of the port from the ship:
To find the bearing of the port from the ship, we can use the inverse tangent function:
Bearing = arctan(vertical component/horizontal component)
Bearing = arctan(-28.98 km/20.37 km)
Bearing ≈ 306.46°
Therefore, the bearing of the port from the ship is approximately 306.46°.
To solve this problem, we can use the concept of vector addition.
(a) To calculate the distance from the port, we need to find the resultant vector of the two given vectors. We can use the Pythagorean theorem to find the magnitude of the resultant vector.
Step 1: Convert the bearings into Cartesian coordinates.
- Bearing 032°: This indicates an angle of 32° measured clockwise from the north direction. In Cartesian coordinates, this can be represented as (21sin(32°), 21cos(32°)).
- Bearing 287°: This indicates an angle of 73° measured clockwise from the north direction. In Cartesian coordinates, this can be represented as (45sin(73°), 45cos(73°)).
Step 2: Add the two vectors component-wise.
- X-components: 21sin(32°) + 45sin(73°)
- Y-components: 21cos(32°) + 45cos(73°)
Step 3: Use the Pythagorean theorem to calculate the magnitude.
- Magnitude: sqrt((X-component)^2 + (Y-component)^2)
(b) To calculate the bearing of the port from the ship, we need to find the angle between the resultant vector and the north direction.
Step 4: Use inverse trigonometric functions to find the bearing.
- Bearing: arctan((Y-component / X-component)) + 90°
Let's perform the calculations:
(a) Distance from the port:
X-component = 21sin(32°) + 45sin(73°) ≈ 37.61 km
Y-component = 21cos(32°) + 45cos(73°) ≈ -18.31 km
Magnitude = sqrt((37.61)^2 + (-18.31)^2) ≈ 41.95 km
Therefore, the ship is approximately 41.95 km from the port.
(b) Bearing of the port from the ship:
Bearing = arctan((-18.31) / (37.61)) + 90° ≈ 129.29°
Therefore, the bearing of the port from the ship is approximately 129.29°.