Question

В прямоугольном треугольнике АВС ( угол С равен 90 градусам )биссектрисы СД и ВЕ пересекаются в точке О. Угол ВОС равен 95 градусов . Найдите острые углы треугольника АВС

Answer 1

Сумма углов треугольника равна 180 
ДСВ= 45 
180-45-95=40= угол ОВС
угол АВС= 2ОВС= 40 *2 = 80
угол BAC= 180 - 90 -80= 10

Answer 2

Given Information:

We are given a right triangle ABC, where angle C is 90 degrees. The bisectors CD and VE intersect at point O. Angle VOC is 95 degrees.

Solution:

To find the acute angles of triangle ABC, we can use the fact that the sum of the angles in a triangle is always 180 degrees.

Let's denote the acute angles of triangle ABC as angle A and angle B.

Since angle C is 90 degrees, we know that angle A + angle B = 90 degrees.

We are also given that angle VOC is 95 degrees. Since angle VOC is formed by the bisectors CD and VE, we can conclude that angle AOC and angle BOV are equal and each is half of angle VOC.

Therefore, angle AOC = angle BOV = 95 degrees / 2 = 47.5 degrees.

Now, we can find angle A and angle B by subtracting angle AOC and angle BOC from 90 degrees.

angle A = 90 degrees - angle AOC = 90 degrees - 47.5 degrees = 42.5 degrees.

angle B = 90 degrees - angle BOC = 90 degrees - 47.5 degrees = 42.5 degrees.

Therefore, the acute angles of triangle ABC are both 42.5 degrees.

Answer: The acute angles of triangle ABC are both 42.5 degrees.