Question

Один из острых углов прямоугольного треугольника = 50 градусов. найдите величину меньшего из углов образованных катетами этого треугольника и медианой. проведенной к гипотенузе

Answer 1

Обозначим треугольник АВС, <B=90° Пусть <ACB=50°
Точку пересечения медианы и гипотенузы обозначим М, то есть медиана ВМ.
 Медиана, проведенная из вершины прямого угла, равна половине гипотенузы, следовательно ВМ=АМ=МС 
Следовательно треугольники АМВ и ВМС равнобедренные. Тогда <CBM=50°, <ABM=40°
Ответ: 40° 

Answer 2

Finding the Angle Formed by the Catheti and the Median

To find the angle formed by the catheti (the two shorter sides) and the median drawn to the hypotenuse of a right triangle, we need to use the properties of right triangles.

Let's denote the acute angle formed by one of the catheti and the hypotenuse as A (which is given as 50 degrees). We want to find the angle formed by the other cathetus and the median drawn to the hypotenuse.

To solve this, we can use the fact that the sum of the angles in a triangle is 180 degrees. Since we already know one angle (50 degrees), we can subtract it from 180 degrees to find the sum of the other two angles.

Let's denote the angle formed by the other cathetus and the median as B. The sum of angles A, B, and the right angle (90 degrees) is equal to 180 degrees.

Therefore, we have the equation:

A + B + 90 = 180

Simplifying the equation, we get:

B = 180 - A - 90

Now we can substitute the given value of angle A (50 degrees) into the equation to find angle B:

B = 180 - 50 - 90

Calculating the value, we find:

B = 40 degrees

So, the angle formed by the smaller cathetus and the median drawn to the hypotenuse is 40 degrees.

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