Answer:
To determine the angles of the triangle defined by the points A, B, and C, we can use the law of cosines. The law of cosines states:
c^2 = a^2 + b^2 - 2ab * cos(θ)
Where:
- c is the length of the side opposite angle θ (in this case, the side opposite the angle we want to find).
- a and b are the lengths of the other two sides of the triangle.
First, we need to find the lengths of the sides of the triangle. We can use the distance formula to calculate the distances between the points A, B, and C:
1. Distance between A and B:
AB = √((x2 - x1)^2 + (y2 - y1)^2)
AB = √((5 - 2)^2 + (8 - 10)^2)
AB = √(3^2 + (-2)^2)
AB = √(9 + 4)
AB = √13
2. Distance between B and C:
BC = √((x2 - x1)^2 + (y2 - y1)^2)
BC = √((4 - 5)^2 + (3 - 8)^2)
BC = √((-1)^2 + (-5)^2)
BC = √(1 + 25)
BC = √26
3. Distance between C and A:
CA = √((x2 - x1)^2 + (y2 - y1)^2)
CA = √((2 - 4)^2 + (10 - 3)^2)
CA = √((-2)^2 + (7)^2)
CA = √(4 + 49)
CA = √53
Now that we have the lengths of all three sides of the triangle, we can use the law of cosines to find the angles θA, θB, and θC:
1. Angle θA (opposite side a = BC):
cos(θA) = (b^2 + c^2 - a^2) / (2 * b * c)
cos(θA) = (AB^2 + CA^2 - BC^2) / (2 * AB * CA)
cos(θA) = (√13^2 + √53^2 - √26^2) / (2 * √13 * √53)
cos(θA) = (13 + 53 - 26) / (2 * √13 * √53)
cos(θA) = 40 / (2 * √13 * √53)
cos(θA) ≈ 0.106
Now, we can find θA by taking the inverse cosine (arccos) of this value:
θA ≈ arccos(0.106) ≈ 1.463 radians (rounded to three decimal places).
2. Angle θB (opposite side b = CA):
cos(θB) = (c^2 + a^2 - b^2) / (2 * c * a)
cos(θB) = (CA^2 + BC^2 - AB^2) / (2 * CA * BC)
cos(θB) = (√53^2 + √26^2 - √13^2) / (2 * √53 * √26)
cos(θB) = (53 + 26 - 13) / (2 * √53 * √26)
cos(θB) = 66 / (2 * √53 * √26)
cos(θB) ≈ 0.301
θB ≈ arccos(0.301) ≈ 1.253 radians (rounded to three decimal places).
3. Angle θC (opposite side c = AB):
cos(θC) = (a^2 + b^2 - c^2) / (2 * a * b)
cos(θC) = (AB^2 + BC^2 - CA^2) / (2 * AB * BC)
cos(θC) = (√13^2 + √26^2 - √53^2) / (2 * √13 * √26)
cos(θC) = (13 + 26 - 53) / (2 * √13 * √26)
cos(θC) = -14 / (2 * √13 * √26)
cos(θC) ≈ -0.301
θC ≈ arccos(-0.301) ≈ 1.893 radians (rounded to three decimal places).
So, the angles of the triangle are approximately:
- θA ≈ 1.463 radians
- θB ≈ 1.253 radians
- θC ≈ 1.893 radians
Explanation: