Question

Окружность с центром в точке о описана около равнобедренного треугольника ABC в котором AB=BC и угол ABC=66 найдите угол BOC . ответ дайте в градусах

Answer 1

См. рисунок.

Есть такая теорема: центральный угол, который опирается на ту же дугу, что и вписанный, в два раза больше его. Значит, AOC=66*2=132°.

Углы BOA и BOC равны, потому что треугольник — равнобедренный. Значит, $BOA=BOC=(360^{\circ}-132^{\circ}):2=228^{\circ}:2=114^{\circ}.$

Answer 2

Problem Analysis

We are given a circle with center O that circumscribes an isosceles triangle ABC, where AB = BC and angle ABC = 66 degrees. We need to find the measure of angle BOC in degrees.

Solution

To find the measure of angle BOC, we can use the fact that the angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the circumference of the circle.

Since triangle ABC is isosceles with AB = BC, we can conclude that angle BAC = angle BCA. Let's denote this angle as x.

Therefore, the measure of angle BOC is twice the measure of angle BAC (or BCA), which is 2x.

To find the value of x, we can use the fact that the sum of the angles in a triangle is 180 degrees. Since triangle ABC is isosceles, we can write the equation:

x + x + 66 = 180

Simplifying the equation, we get:

2x + 66 = 180

Subtracting 66 from both sides, we have:

2x = 180 - 66 2x = 114

Dividing both sides by 2, we find:

x = 57

Therefore, the measure of angle BOC is twice the measure of angle BAC (or BCA), which is 2x = 2 * 57 = 114 degrees.

Answer: The measure of angle BOC is 114 degrees.