given the points A(0,0), B(3,1) and C (1,4)

what is the measure of angle ABC
I plotted it on a gragh and got
1.46 degress. I am unsure if I am doing this question correctly. Or am I misreading it
thanks in advance .

A simple sketch will show you that that answer cannot possibly be right

Extend CB to cut the x-axis at D
(using the basic concept that if the slope of a line is m and the line forms an angle of Ø with the x-axis
then tanØ = m)

slope of AB = 1/3
so tan BAD = 1/3
BAD = 18.43°
slope of CB = -3/2 , angle is obtuse
tan (angle of line with x-axis) = 123.69°
so angle BDA = 56.304°
and angle ABC is the exterior angle
and angle ABC = angle BAD + angle BDA
= 18.43° + 56.304° = 74.74°

or

If you know vectors
vector CB = (-2,3)
vector BA = (-3,-1)

v(CB) ∙ v(BA) = |CB||BA)cosØ, where Ø is our angle
6 - 3 = √13 √10 cos‚
cosØ = 3/√130 = .26311..
Ø = 74.74° as above.

To find the measure of angle ABC, you can use the distance formula and the cosine rule.

First, let's find the lengths of the sides AB, AC, and BC:

AB = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(3 - 0)^2 + (1 - 0)^2]
= √[9 + 1]
= √10

AC = √[(x3 - x1)^2 + (y3 - y1)^2]
= √[(1 - 0)^2 + (4 - 0)^2]
= √[1 + 16]
= √17

BC = √[(x3 - x2)^2 + (y3 - y2)^2]
= √[(1 - 3)^2 + (4 - 1)^2]
= √[4 + 9]
= √13

Next, we can use the cosine rule to find the measure of angle ABC:

cos(ABC) = (AB^2 + BC^2 - AC^2) / (2 * AB * BC)
= (√10^2 + √13^2 - √17^2) / (2 * √10 * √13)
= (10 + 13 - 17) / (2 * √10 * √13)
= 6 / (2 * √10 * √13)
= 3 / (√10 * √13)
= 3 / √130

To find the measure of angle ABC, we need to find the arccosine of this value:

angle ABC = arccos(3 / √130)
≈ 48.03 degrees

Therefore, the measure of angle ABC is approximately 48.03 degrees, which is different from your initial calculation of 1.46 degrees. Please double-check your calculations and ensure that you have correctly entered the coordinates.

To find the measure of angle ABC, you can use the concept of slopes and the distance formula between the points.

First, calculate the slopes of lines AB and BC. The slope of a line passing through two points (x1, y1) and (x2, y2) can be calculated using the formula:
m = (y2 - y1) / (x2 - x1)

For line AB:
m1 = (1 - 0) / (3 - 0) = 1/3

For line BC:
m2 = (4 - 1) / (1 - 3) = 3/2

The measure of angle ABC can be calculated using the formula:
tan(θ) = |(m2 - m1) / (1 + m1 * m2)|

Substituting the values:
tan(θ) = |((3/2) - (1/3)) / (1 + (1/3) * (3/2))|

Simplifying the expression:
tan(θ) = |(6/6 - 2/6) / (6/6 + 1/6)| = |4/6| = 2/3

Now, you need to find the inverse tangent of 2/3 to get the angle measure.
θ = atan(2/3) ≈ 33.69 degrees

Based on your statement that you obtained an angle measure of 1.46 degrees, it seems there might be an error in either the calculations or the interpretation of the plot. Check your calculations again and make sure you are inputting the correct coordinates. It's always a good idea to verify the answer using multiple methods, such as calculating the angle using trigonometric functions or vector operations.