Question

На сторонах равнобедренного треугольника ABC с основанием AC отмечены точки E и D так, что BE=ED и AD=AC. Угол при вершине B равен 30 градусам. Найдите угол EDA ПОМОГИТЕ

Answer 1

Ответ:75°

Объяснение:

Answer 2

Given Information

We are given a triangle ABC with base AC. Points E and D are marked on the sides of the isosceles triangle ABC such that BE = ED and AD = AC. The angle at vertex B is 30 degrees. We need to find the angle EDA.

Solution

To find the angle EDA, we can use the fact that the sum of the angles in a triangle is 180 degrees. Let's break down the problem step by step.

1. Triangle ABC is an isosceles triangle, which means that AB = BC. Since AD = AC, we can conclude that AD = AB + BC. 2. Let's assume that AB = x. Then, BC = x as well. Since AD = AC, we have AD = x + x = 2x. 3. We know that the angle at vertex B is 30 degrees. Since triangle ABC is isosceles, the angles at vertices A and C are also equal. Let's call this angle α. 4. The sum of the angles in triangle ABC is 180 degrees. Therefore, we have: - Angle A + Angle B + Angle C = 180 degrees - α + 30 + α = 180 - 2α + 30 = 180 - 2α = 150 - α = 75 5. Now that we know the value of α, we can find the angles EDA and EAD. Since triangle AED is isosceles (BE = ED), the angles EDA and EAD are equal. Let's call these angles β. 6. The sum of the angles in triangle AED is 180 degrees. Therefore, we have: - Angle EDA + Angle EAD + Angle AED = 180 degrees - β + β + α = 180 - 2β + 75 = 180 - 2β = 105 - β = 52.5 7. Therefore, the angle EDA is 52.5 degrees.

Answer

The angle EDA is 52.5 degrees.