Question

В ромбе ABCD диогонали пересекаются в точке O угол ADC = 108°Найти углы треугольника AOB с дано и решением и желательно с фото

Answer 1

Фото не загружается. Дано: абсд - ромб; <адс=108°; ас пересекает бд в точке О; бд перпендикулярно ас. Найти: углы треугольника або. Решение: 1. <адс= <абс (в ромбе противолежащие углы равны.) бд - биссектриса <абс |=> 108:2=54°.(<або). 2. <боа=90°(т.к. диагонали перпендикулярны.) 3. 54+90=144 180-144=36°(<оаб). Ответ: <або=54°; <боа=90°; <оаб=36°

Answer 2

Finding the angles of triangle AOB in a rhombus

To find the angles of triangle AOB in a rhombus ABCD, where the diagonals intersect at point O and angle ADC is 108°, we can use the properties of a rhombus and the relationships between the angles in a triangle.

In a rhombus, opposite angles are equal. Therefore, angle AOB is equal to angle COD.

Let's denote angle AOB as x. Since angle ADC is 108°, angle COD is also 108°.

Now, let's consider triangle AOB. The sum of the angles in a triangle is always 180°. Therefore, we can write the equation:

x + x + 108° = 180°

Simplifying the equation:

2x + 108° = 180°

Subtracting 108° from both sides:

2x = 72°

Dividing both sides by 2:

x = 36°

So, angle AOB is 36°.

Therefore, the angles of triangle AOB in the rhombus are: - Angle AOB = 36°

Please note that the solution provided is based on the given information and the properties of a rhombus. However, without a diagram or additional information, it is not possible to provide a visual representation or further details.

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