Question

Даны вершины треугольника A(1;1), B(4;1), C(4;5) вычислите косинусы его угловСрочно очень срочно!!!!! Прошу!!!!!

Answer 1

Треугольник  АВС прямоугольный, уголВ=90, АВ=4-1=3, СВ=5-1=4, АС=корень(АВ в квадрате+СВ в квадрате)=корень(9+16)=5,
cosA =AB/AC=3/5=0.6, cosC=CB/AC=4/5=0.8, cosB (90)=0

Answer 2

Находим вектора AB=(3;0) и их длины |AB|=3
                             AC=(3;4)                    |AC|=5
                            cosA=(3*3+4*0)/3*5=0.6
                             CB=(0;-4)  |CB|=4
                             cosC=16/4*5=0,8
cosB=(-3*0+4*0)/3*4=0

Answer 3

Calculating the Cosines of the Angles of a Triangle

To calculate the cosines of the angles of a triangle, we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides of lengths a, b, and c, and opposite angles A, B, and C, the following equation holds:

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

In this case, we are given the coordinates of the vertices of the triangle: A(1,1), B(4,1), and C(4,5). To calculate the cosines of the angles, we need to find the lengths of the sides of the triangle.

Calculating the Lengths of the Sides

To find the lengths of the sides of the triangle, we can use the distance formula. The distance formula states that the distance between two points (x1, y1) and (x2, y2) is given by:

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

Using this formula, we can calculate the lengths of the sides AB, BC, and AC.

AB = sqrt((4 - 1)^2 + (1 - 1)^2) = sqrt(9) = 3 BC = sqrt((4 - 4)^2 + (5 - 1)^2) = sqrt(16) = 4 AC = sqrt((4 - 1)^2 + (5 - 1)^2) = sqrt(25) = 5

Calculating the Cosines of the Angles

Now that we have the lengths of the sides, we can use the Law of Cosines to calculate the cosines of the angles. Let's calculate the cosine of angle A first.

cos(A) = (b^2 + c^2 - a^2) / (2bc) = (4^2 + 5^2 - 3^2) / (2 * 4 * 5) = (16 + 25 - 9) / 40 = 32 / 40 = 0.8

Similarly, we can calculate the cosines of angles B and C.

cos(B) = (c^2 + a^2 - b^2) / (2ca) = (5^2 + 3^2 - 4^2) / (2 * 5 * 3) = (25 + 9 - 16) / 30 = 18 / 30 = 0.6

cos(C) = (a^2 + b^2 - c^2) / (2ab) = (3^2 + 4^2 - 5^2) / (2 * 3 * 4) = (9 + 16 - 25) / 24 = 0 / 24 = 0

Therefore, the cosines of the angles of the triangle ABC are: - cos(A) = 0.8 - cos(B) = 0.6 - cos(C) = 0

Please note that the angles themselves can be calculated using the inverse cosine function (arccos) if needed.