Angela traveled a distance of 42km from A, on a bearing of 080 degree to B. At B she then traveled a further distance of 20km to C on the bearing of 160 degree. Calculate a) distance/AC/
b) the bearing of A from C.
See ken's post, and come on back with some of your ideas.
Always start by drawing the diagram
recall how to convert polar form to x-y coordinates for adding up distances
and study up on law of cosines
...... and please note that you take a bearing on something to find its direction from you.
HOWEVER if you sail for it, you travel on a HEADING not a BEARING.
All angles are measured CW from +y-axis.
AC = AB + BC = 42km[80o] + 20km[160o].
X = 42*sin80 + 20*sin160 = 48.2 km.
Y = 42*Cos80 + 20*Cos160 = -11.5 km.
a. AC = sqrt ( X^2+Y^2).
b. TanA = X/Y.
To solve this problem, we can use trigonometry and vector addition. To make it easier, we will break down the journey into two segments: AB and BC.
a) To find the distance AC, we need to calculate the magnitude of the vector AC. We can use the cosine rule to do this.
1. Find the angle BAC:
The angle BAC can be calculated by subtracting the bearing of B from the bearing of A.
Bearing of B = 080 degrees
Bearing of A = 180 degrees - 80 degrees = 100 degrees
Angle BAC = 100 degrees - 80 degrees = 20 degrees
2. Use the cosine rule:
The cosine rule states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of the two sides multiplied by the cosine of the angle between them.
Let's denote the sides of the triangle ABC as a, b, and c, and the angle BAC as θ.
Using the cosine rule:
AC^2 = AB^2 + BC^2 - 2 * AB * BC * cos(θ)
AC^2 = 42^2 + 20^2 - 2 * 42 * 20 * cos(20 degrees)
AC^2 = 1764 + 400 - 1680 * 0.94
AC^2 = 1764 + 400 - 1581.6
AC^2 = 582.4
Taking the square root of both sides:
AC = √582.4
AC ≈ 24.11 km
Therefore, the distance AC is approximately 24.11 km.
b) To find the bearing of A from C, we can use vector addition.
Imagine A, B, and C as points in a coordinate system.
1. Calculate the vector AB:
To do this, we can represent AB as a displacement vector. The horizontal component (ABx) can be obtained by multiplying the distance AB (42 km) by the cosine of the bearing (080 degrees). The vertical component (ABy) can be obtained by multiplying the distance AB by the sine of the bearing.
ABx = AB * cos(080 degrees)
ABx = 42 * cos(080 degrees)
ABx ≈ -40.87 km
ABy = AB * sin(080 degrees)
ABy = 42 * sin(080 degrees)
ABy ≈ 8.91 km
So, AB ≈ (-40.87 km, 8.91 km)
2. Calculate the vector BC:
Using a similar process, we can calculate BC. The horizontal component (BCx) is given by the distance BC multiplied by the cosine of the bearing (160 degrees), while the vertical component (BCy) is given by the distance BC multiplied by the sine of the bearing.
BCx = BC * cos(160 degrees)
BCx = 20 * cos(160 degrees)
BCx ≈ -5.59 km
BCy = BC * sin(160 degrees)
BCy = 20 * sin(160 degrees)
BCy ≈ 19.17 km
So, BC ≈ (-5.59 km, 19.17 km)
3. Calculate the vector AC:
To find the vector AC, we can add vectors AB and BC by summing their horizontal and vertical components separately.
ACx = ABx + BCx
ACx = -40.87 km + (-5.59 km)
ACx ≈ -46.46 km
ACy = ABy + BCy
ACy = 8.91 km + 19.17 km
ACy ≈ 28.08 km
So, AC ≈ (-46.46 km, 28.08 km)
4. Calculate the bearing of A from C:
The bearing from one point to another can be calculated as the angle formed between the vector connecting the two points and the positive x-axis.
Using trigonometry:
θ = tan^(-1)(ACy/ACx)
θ = tan^(-1)(28.08 km / -46.46 km)
Evaluating this in a calculator, we find:
θ ≈ -30.38 degrees
However, bearings are usually measured clockwise from the north direction. To convert this angle into a bearing, we need to add 360 degrees.
Bearing of A from C = 360 degrees - 30.38 degrees
Bearing of A from C ≈ 329.62 degrees
Therefore, the bearing of A from C is approximately 329.62 degrees.