Question

В треугольнике ABC угол С равен 90 градусовсем, синус внешнего угла при вершине А равен 3/5 ,АВ = 10 . Найдите АС . Помогите, кто может

Answer 1

Если синус внешнего угла при вершине А равен 3/5, значит и синус смежного с ним угла тоже равен 3/5, то есть
A
                                         Sin< A = 3/5
                                         Из прямоугольного ΔABC  Sin<A = BC/AB ,
                                         отсюда BC = AB * Sin<A = 10 * 3/5 = 6
C                   B                  По теореме Пифагора
                                          AC² = AB² - BC² = 100 - 36 = 64     AC = √64 = 8

Answer 2

Problem Analysis

We are given a right triangle ABC, where angle C is 90 degrees. We are also given that the sine of the external angle at vertex A is 3/5, and AB is equal to 10. We need to find AC.

Solution

To solve this problem, we can use the trigonometric relationship between the sine of an angle and the lengths of the sides of a right triangle.

The sine of an angle in a right triangle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. In this case, the sine of the external angle at vertex A is given as 3/5.

Let's denote the length of AC as x. According to the given information, AB is equal to 10.

Using the sine ratio, we can write the equation:

sin(A) = AC / AB

Substituting the given values, we have:

3/5 = x / 10

To solve for x, we can cross-multiply:

3 * 10 = 5 * x

Simplifying further:

30 = 5x

Dividing both sides by 5:

6 = x

Therefore, the length of AC is 6.

Answer

The length of AC is 6.