Find, correct to the nearest degree, the three angles of the triangle with the given vertices.
A(0, 1, 3), B(-2, 4, 5), C(1, 4, -3),
∠CAB = °
∠ABC = °
∠BCA = °
http://math.stackexchange.com/questions/361412/finding-the-angle-between-three-points
To find the angles of the triangle, we need to calculate the lengths of the three sides first.
Let's find the lengths of the sides AB, BC, and AC.
The distance between two points in 3D space can be calculated using the distance formula:
d = √[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]
For AB:
dAB = √[(-2 - 0)^2 + (4 - 1)^2 + (5 - 3)^2]
= √[4 + 9 + 4]
= √17
For BC:
dBC = √[(1 - -2)^2 + (4 - 4)^2 + (-3 - 5)^2]
= √[9 + 64 + 64]
= √137
For AC:
dAC = √[(1 - 0)^2 + (4 - 1)^2 + (-3 - 3)^2]
= √[1 + 9 + 36]
= √46
Now that we have the lengths of each side, we can use the law of cosines to find the angles.
The law of cosines states:
cos(A) = (b² + c² - a²) / (2bc)
For ∠CAB:
cos(CAB) = (dBC² + dAC² - dAB²) / (2 * dBC * dAC)
= (137² + 46² - 17²) / (2 * 137 * 46)
= (18769 + 2116 - 289) / (2 * 137 * 46)
= 21096 / (2 * 137 * 46)
≈ 0.384
∠CAB = arccos(0.384)
≈ 67.5° (rounded to the nearest degree)
For ∠ABC:
cos(ABC) = (dAB² + dBC² - dAC²) / (2 * dAB * dBC)
= (17² + 137² - 46²) / (2 * 17 * 137)
= (289 + 18769 - 2116) / (2 * 17 * 137)
= 15842 / (2 * 17 * 137)
≈ 0.716
∠ABC = arccos(0.716)
≈ 43.6° (rounded to the nearest degree)
For ∠BCA:
cos(BCA) = (dAC² + dBC² - dAB²) / (2 * dAC * dBC)
= (46² + 137² - 17²) / (2 * 46 * 137)
= (2116 + 18769 - 289) / (2 * 46 * 137)
= 21096 / (2 * 46 * 137)
≈ 0.384
∠BCA = arccos(0.384)
≈ 67.5° (rounded to the nearest degree)
Therefore, the three angles of the triangle with the given vertices are approximately:
∠CAB ≈ 67.5°
∠ABC ≈ 43.6°
∠BCA ≈ 67.5°
To find the three angles of a triangle with given vertices, we can use the dot product formula:
cos(angle) = (AB · AC) / (|AB| * |AC|)
where AB and AC are the vectors formed by subtracting the coordinates of the vertices.
Let's find the angles step by step:
1. Calculate the vectors AB and AC:
AB = (Bx - Ax, By - Ay, Bz - Az)
= (-2 - 0, 4 - 1, 5 - 3)
= (-2, 3, 2)
AC = (Cx - Ax, Cy - Ay, Cz - Az)
= (1 - 0, 4 - 1, -3 - 3)
= (1, 3, -6)
2. Calculate the dot products AB · AC:
AB · AC = ABx * ACx + ABy * ACy + ABz * ACz
= (-2 * 1) + (3 * 3) + (2 * -6)
= -2 + 9 - 12
= -5
3. Calculate the magnitudes |AB| and |AC|:
|AB| = sqrt(ABx^2 + ABy^2 + ABz^2)
= sqrt((-2)^2 + 3^2 + 2^2)
= sqrt(4 + 9 + 4)
= sqrt(17)
|AC| = sqrt(ACx^2 + ACy^2 + ACz^2)
= sqrt(1^2 + 3^2 + (-6)^2)
= sqrt(1 + 9 + 36)
= sqrt(46)
4. Calculate the angles using the cos(angle) formula:
∠CAB = arccos((-5) / (sqrt(17) * sqrt(46)))
= arccos(-5 / (sqrt(782)))
≈ arccos(-0.564)
≈ 127.281 degrees (rounded to the nearest degree)
∠ABC = arccos((-5) / (sqrt(17) * sqrt(46)))
= arccos(-5 / (sqrt(782)))
≈ arccos(-0.564)
≈ 127.281 degrees (rounded to the nearest degree)
∠BCA = arccos((-5) / (sqrt(17) * sqrt(46)))
= arccos(-5 / (sqrt(782)))
≈ arccos(-0.564)
≈ 127.281 degrees (rounded to the nearest degree)
Therefore, ∠CAB ≈ 127 degrees, ∠ABC ≈ 127 degrees, and ∠BCA ≈ 127 degrees.