Question

Найти угол B в треугольнике ABC , если известны координаты его вершин A(4; 8), B(6; 9), C(2; 12).

Answer 1

Даны точки A(4; 8), B(6; 9), C(2; 12) как вершины треугольника.

Найти угол треугольника можно двумя способами:

-1)  векторным,

-2) по теореме косинусов.

1) Вектор ВА = (4-6; 8-9) = (-2; -1), его модуль (длина) |AB| = √((-2)² + (-1)²) = √5.

  Вектор ВС = (2-6; 12-9) = (-4; 3), его модуль (длина) |AC| = √((-4)² + 3²) = √25 = 5.

cos(ВА_ВC) = ((-2)*(-4) + (-1)*3)/(√5*5) = 5/(5*√5) = √5/5 ≈ 0,4472.

B = arccos 0,44721 = 1,1072 радиан или 63,435 градуса.

2) По разности координат находим длины сторон.

Расчет длин сторон  

АВ (с) = √((Хв-Ха)²+(Ув-Уа)²) = √5 ≈ 2,2361.

BC (а)= √((Хc-Хв)²+(Ус-Ув)²) = √25 = 5.

AC (в) = √((Хc-Хa)²+(Ус-Уa)²) = √20 ≈ 4,4721.

Находим косинус угла В.

cos В = (5 + 25 - 20) / (2*5*√5) = 10/(2*5*√5) = 1/√5 = 0,4472.

B = arccos 0,44721 = 1,1072 радиан или 63,435 градуса.

Answer 2

Finding the Angle B in Triangle ABC

To find the angle B in triangle ABC, we can use the coordinates of its vertices A(4, 8), B(6, 9), and C(2, 12). One way to calculate the angle is by using the Law of Cosines.

The Law of Cosines states that in a triangle with sides a, b, and c, and angle C opposite side c, the following equation holds:

c^2 = a^2 + b^2 - 2ab * cos(C)

In our case, we want to find angle B, so we can use the sides AB, BC, and angle B. Let's label the sides as follows: - Side AB: a - Side BC: b - Side AC: c

Now, let's calculate the lengths of the sides AB, BC, and AC using the distance formula:

AB = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Using the coordinates of points A(4, 8) and B(6, 9), we can calculate AB as follows:

AB = sqrt((6 - 4)^2 + (9 - 8)^2) = sqrt(2^2 + 1^2) = sqrt(5)

Similarly, using the coordinates of points B(6, 9) and C(2, 12), we can calculate BC as follows:

BC = sqrt((2 - 6)^2 + (12 - 9)^2) = sqrt((-4)^2 + 3^2) = sqrt(25) = 5

Now, we can substitute the values of AB, BC, and AC into the Law of Cosines equation to find angle B:

BC^2 = AB^2 + AC^2 - 2 * AB * AC * cos(B)

Substituting the values we calculated earlier:

5^2 = sqrt(5)^2 + AC^2 - 2 * sqrt(5) * AC * cos(B)

Simplifying the equation:

25 = 5 + AC^2 - 2 * sqrt(5) * AC * cos(B)

20 = AC^2 - 2 * sqrt(5) * AC * cos(B)

Now, we need to find the value of AC. Using the coordinates of points A(4, 8) and C(2, 12), we can calculate AC as follows:

AC = sqrt((2 - 4)^2 + (12 - 8)^2) = sqrt((-2)^2 + 4^2) = sqrt(20) = 2 * sqrt(5)

Substituting the value of AC into the equation:

20 = (2 * sqrt(5))^2 - 2 * sqrt(5) * 2 * sqrt(5) * cos(B)

Simplifying further:

20 = 20 - 20 * cos(B)

Cancelling out the common factor of 20:

1 = -cos(B)

To find the value of cos(B), we can take the inverse cosine (arccos) of both sides:

B = arccos(-1)

The arccosine of -1 is equal to 180 degrees or π radians.

Therefore, the angle B in triangle ABC is 180 degrees or π radians.

Please note that the calculations provided above are based on the given coordinates and the assumption that the triangle ABC is a non-degenerate triangle.