Given 3 points, find the angle?

Question

Given three points, (Xa,Ya), (Xb,Yb), and (Xc,Yc), find angle ABC.

Answer 1

Let A(Xa,Ya),B(Xb,Yb),C(Xc,Yc)

We will be using cosine rule.
AC*AC = AB*AB + BC*BC - 2*AB*AC*cos ABC
cos ABC = (AB*AB + BC*BC - AC*AC) / (2*AB*AC)
ABC = arc cos (AB*AB + BC*BC - AC*AC) / (2*AB*AC)

where:
AC is the distance between point A and C.
AB is the distance between point A and B.
BC is the distance between point B and C.

Hope it helps.
peace
vixklen

Answer 2

This is too tedious to type on regular QWERTY keyboard.

However, to find the angle ABC ( i.e. the vertex is B) or angle B, 1) use SINE LAW, viz.

a/sinA = b/sinB = c/sinC

a, b, and c are lengths of the sides of the triangle formed by the three points.

2) Then, use the the distance formula to get a, b, and c.

For example, side b = square root of( (Xb-Xa)squared + (Yb-Ya)squared))

Sorry, the yahoo page does not have options for math symbols.

3) To get the angle B, take the arcsin of B from equation 1)

Answer 3

(Xc - Xa)^2 + (Yc + Ya)^2 = (Xa - Xb)^2 +(Ya - Yb)^2 + (Xc - Xb)^2 + (Yc - Yb)^2 - 2sqrt((Xa - Xb)^2 +(Ya - Yb)^2)((Xc - Xb)^2 + (Yc - Yb)^2))cos(ABC)

(Xc - Xa)^2 + (Yc + Ya)^2 + (Xa - Xb)^2 +(Ya - Yb)^2 + (Xc - Xb)^2 + (Yc - Yb)^2 - (Xc - Xa)^2 + (Yc + Ya)^2 = 2sqrt((Xa - Xb)^2 +(Ya - Yb)^2)((Xc - Xb)^2 + (Yc - Yb)^2))cos(ABC)

cos(ABC) = ((Xa - Xb)^2 +(Ya - Yb)^2 + (Xc - Xb)^2 + (Yc - Yb)^2 - (Xc - Xa)^2 + (Yc - Ya)^2) / 2sqrt((Xa - Xb)^2 +(Ya - Yb)^2)((Xc - Xb)^2 + (Yc - Yb)^2))

cos(ABC) = (- XaXb + Xb^2 - YaYb + Yb^2 - XbXc - YbYc + XaXc + YaYc) / sqrt(Xa^2 - 2XaXb + Xb^2 + Ya^2 - 2YaYb + Yb^2)(Xc^2 - 2XbXc + Xb^2 + Yc^2 - 2YbYc + Yb^2))

Answer 4

There are a pair of distinctive procedures you will be able to desire to try this. one way is to apply the gap formula to locate the lengths of AB, BC, and AC. Then to locate the perspective at B, use the regulation of Cosines: (AC)^2 = (AB)^2 + (BC)^2 - 2(AB)(BC)cos(ABC) Plug interior the values you get for AC, AB, and BC, then resolve this for cos(ABC). Then take the inverse cosine to locate the degree of perspective ABC.

Answer 5

given the three points first solve the distance between the three point so you can use it the length of the triangles

then solve the angle difference between their tangents

Answer 6

Find the vectors from B to A and B to C:

u=
v=

Then find the angle between the two vectors:

θ = arccos (u·v/(||u||*||v||)

Alternatively:

find the slope of the two lines:

M_1=(y_a-y_b)/(x_a-x_b)
M_2=(y_c-x_b)/(x_c-x_b)

Then find the difference of the angles of the two slopes:

θ=| arctan M_1 - arctan M_2 |

Answer 7

join those three points. Find the distance between 1st &2nd,2nd &3rd ,3rd &1st using distance formula.then find the angle using t-ratios

Answer 8

180-a = b+c