Question

Угол А остроугольного треугольника =50 градусов.Найдите тупой угол ,под которым пересекаются биссектриса угла А и высота,опущенная из вершины В.Ответ 115,но нужен рисунок. Пжл срочно ребят!!!!!!!!!!

Answer 1

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Answer 2

Finding the obtuse angle formed by the bisector of angle A and the altitude from vertex B in an acute triangle

To find the obtuse angle formed by the bisector of angle A and the altitude from vertex B in an acute triangle, we need to consider the properties of acute triangles and the relationships between angles and sides.

Let's start by drawing a diagram of the triangle. We have an acute triangle ABC, where angle A is 50 degrees. We want to find the obtuse angle formed by the bisector of angle A and the altitude from vertex B.

       B
      / \
     /   \
    /     \
   /       \
  /         \
 /___________\
A      C

To find the obtuse angle, we need to determine the measure of angle BAC and angle ABC.

Since angle A is 50 degrees, we know that angle BAC is also 50 degrees because it is an isosceles triangle (AB = AC). This is because the bisector of angle A divides the opposite side (BC) into two equal parts.

Now, let's find angle ABC. Since angle BAC is 50 degrees, angle ABC is equal to 180 degrees minus the sum of angles BAC and BCA. In this case, angle BCA is 90 degrees because it is the right angle formed by the altitude from vertex B. Therefore, angle ABC is:

angle ABC = 180 degrees - (angle BAC + angle BCA) = 180 degrees - (50 degrees + 90 degrees) = 180 degrees - 140 degrees = 40 degrees

So, the obtuse angle formed by the bisector of angle A and the altitude from vertex B is 40 degrees.

Please note that the answer you provided (115 degrees) does not match the calculations based on the given information. It is important to double-check the calculations and ensure the accuracy of the answer.

I hope this explanation helps! Let me know if you have any further questions.