Question

На карте, масштаб которой 1:2 000 000, расстояние между пунктами А и В равно 5√3 см, а расстояние между пунктами А и С равно 5 см. Найдите расстояние между пунктами В и С на местности, если на карте уголACB = 120° ПОЖАЛУЙСТА ЭТО ОЧЕНЬ СРОЧНО ​

Answer 1

Ответ:

Ну, как-нибудь треугольник изобразишь самостоятельно.

Объяснение:

Из теоремы косинусов, приняв неизвестную сторону за x:

cos(120) = (25 + x² - 75) / 10*x; cos(120) = -0.5

x² + 5*x - 50 = 0;

x = (-10; 5). Треугольник равнобедренный.

Пропорция? Зачем? Умножаем на знаменатель масштаба и все.

5 см * 2000000 = 10000000 см = 100 км

Answer 2

Calculating Distance on the Map

To calculate the distance between points B and C on the map, we can use the given information about the distances between points A and B, and points A and C, as well as the angle ACB.

Given: - Scale of the map: 1:2,000,000 - Distance between points A and B: 5√3 cm - Distance between points A and C: 5 cm - Angle ACB: 120°

We can use the law of cosines to find the distance between points B and C on the map.

The law of cosines states that for any triangle with sides of length a, b, and c, and the angle opposite side c being C, the following equation holds true: c^2 = a^2 + b^2 - 2ab * cos(C)

Let's calculate the distance between points B and C on the map using the given information.

Calculating the Distance

Using the law of cosines, we can calculate the distance between points B and C on the map:

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

Where: - a = Distance between points A and B - b = Distance between points A and C - C = Angle ACB

Substituting the given values: c^2 = (5√3)^2 + 5^2 - 2 * 5√3 * 5 * cos(120°)

Now, let's calculate the distance between points B and C on the map.

Calculation

c^2 = (75) + 25 - 50√3 * (-1/2) c^2 = 75 + 25 + 25√3 c^2 = 100 + 25√3 c = √(100 + 25√3)

Therefore, the distance between points B and C on the map is approximately √(100 + 25√3) cm.

This calculation provides the distance between points B and C on the map based on the given scale, distances, and angle.