2) In the given figure, angle ABC = 40 degrees. Find angle x.

The angle subtended by an arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle.

In the given figure, $$\angle OBC = \angle OCB$$ since $$OB = OC$$ (radii).

Therefore, $$\angle OBC = \angle OCB = \frac{180^{\circ} - 40^{\circ}}{2} = \frac{140^{\circ}}{2} = 70^{\circ}$$.

However, this is incorrect as angle ABC is given as 40 degrees and it seems to be an angle formed by a tangent and a chord.

Let's assume the line passing through B and C is a tangent to the circle at point B. Then, the angle between the tangent and a chord through the point of contact is equal to the angle in the alternate segment.

If we assume that BC is a chord and the line is tangent at B, then angle ABC would be related to the angle subtended by chord BC in the alternate segment.

Let's reconsider the diagram. It seems that O is the center of the circle. AB is a chord, and BC is a tangent at B. Angle ABC is given as 40 degrees. Angle BOC is the angle subtended by arc BC at the center. Angle BAC is the angle subtended by arc BC at the circumference.

There is an angle labeled 'x' which is $$\angle BOC$$. The angle labeled '40' is $$\angle ABC$$. The line segment AB is a chord and BC is a tangent at B.

According to the alternate segment theorem, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

So, $$\angle ABC = \angle BAC = 40^{\circ}$$.

In triangle ABO, OA = OB (radii), so it is an isosceles triangle. We need to find $$\angle BOC$$, which is denoted by x.

Let's assume the angle marked 40 degrees is $$\angle BAC$$ and the line segment BC is a chord. Then the angle x is $$\angle BOC$$. In this case, $$\angle BOC = 2 * \angle BAC = 2 * 40^{\circ} = 80^{\circ}$$. This does not fit the diagram as BC is shown as a tangent.

Let's assume that the line passing through A and B is a tangent at A, and BC is a chord. This also does not fit the diagram.

Let's assume O is the center. AB is a chord and BC is a tangent at B. The angle marked 40 degrees is $$\angle ABO$$. If $$\angle ABO = 40^{\circ}$$, then since OA=OB, $$\angle OAB = 40^{\circ}$$. Then $$\angle AOB = 180 - 40 - 40 = 100^{\circ}$$.

Let's assume the 40 degrees is $$\angle OAB$$. Then $$\angle OBA = 40^{\circ}$$ and $$\angle AOB = 100^{\circ}$$.

Let's assume the 40 degrees is $$\angle OBC$$. Since OB is radius and BC is tangent, $$\angle OBC = 90^{\circ}$$. This is not possible.

Let's assume the 40 degrees is $$\angle BAC$$. Then $$\angle BOC = 2 * \angle BAC = 80^{\circ}$$.

Let's assume the 40 degrees is $$\angle BCA$$. Then in triangle ABC, if BC is a chord, we don't know enough.

Given the diagram, the most likely interpretation is that BC is tangent to the circle at B, and $$\angle ABC = 40^{\circ}$$. AB is a chord. O is the center. x is $$\angle BOC$$. The angle labeled 40 is $$\angle ABC$$.

By the alternate segment theorem, the angle between the tangent BC and the chord AB is equal to the angle subtended by the chord AB in the alternate segment, which is $$\angle AOB$$. No, it is $$\angle ACB$$. Oh, the angle 'x' is $$\angle BOC$$.

Let's assume $$\angle CAB = 40^{\circ}$$. Then $$\angle COB = 2 * \angle CAB = 80^{\circ}$$. This is likely if BC is a chord.

If BC is a tangent at B, and AB is a chord, and $$\angle ABC = 40^{\circ}$$. Then by alternate segment theorem, $$\angle BAC = \angle ABC = 40^{\circ}$$.

In triangle ABO, OA = OB. So, $$\angle OAB = \angle OBA$$. Let's denote this angle as y.

We are given $$\angle ABC = 40^{\circ}$$. If BC is tangent at B, then $$\angle OBC = 90^{\circ}$$. This contradicts the diagram.

Let's assume that the angle 40 degrees is $$\angle BAC$$. Then the angle subtended by arc BC at the center is $$\angle BOC = x$$. The angle subtended by arc BC at the circumference is $$\angle BAC$$. So, $$x = 2 * \angle BAC = 2 * 40^{\circ} = 80^{\circ}$$.

Let's assume that the angle 40 degrees is $$\angle OAB$$. Then since OA=OB, $$\angle OBA = 40^{\circ}$$. Then $$\angle AOB = 180 - 40 - 40 = 100^{\circ}$$.

Let's assume that the angle 40 degrees is $$\angle OBA$$. Then since OA=OB, $$\angle OAB = 40^{\circ}$$. Then $$\angle AOB = 180 - 40 - 40 = 100^{\circ}$$.

Let's assume that the angle 40 degrees is $$\angle ACB$$. Then $$\angle AOB = 2 * \angle ACB = 80^{\circ}$$.

Let's assume that the angle marked 40 degrees is $$\angle ABC$$, and BC is a tangent at B. Then the alternate segment theorem states that the angle between the tangent BC and the chord AB is equal to the angle subtended by the chord AB in the alternate segment. That angle is $$\angle ACB$$. So $$\angle ACB = 40^{\circ}$$.

In triangle ABC, we have $$\angle ABC = 40^{\circ}$$ and $$\angle ACB = 40^{\circ}$$. Thus, triangle ABC is isosceles with AB = AC. This does not seem to help find x which is $$\angle BOC$$.

Let's assume the 40 degrees is $$\angle BAC$$. Then $$\angle BOC = x = 2 * \angle BAC = 2 * 40^{\circ} = 80^{\circ}$$.

Let's assume the 40 degrees is $$\angle OAB$$. Since OA=OB, $$\angle OBA = 40^{\circ}$$. Then $$\angle AOB = 180 - 80 = 100^{\circ}$$.

Let's assume the 40 degrees is $$\angle OBA$$. Since OA=OB, $$\angle OAB = 40^{\circ}$$. Then $$\angle AOB = 180 - 80 = 100^{\circ}$$.

Let's assume that the angle $$40^{\circ}$$ is $$\angle BAC$$. Then the angle subtended by arc BC at the center is $$x = \angle BOC = 2 * \angle BAC = 2 * 40^{\circ} = 80^{\circ}$$.

If the 40 degrees is $$\angle ABC$$ and BC is a tangent at B. Then by alternate segment theorem, the angle between the tangent BC and chord AB is equal to the angle in the alternate segment, i.e., $$\angle ACB = 40^{\circ}$$.

In $$\triangle ABO$$, OA = OB (radii). Let $$\angle OAB = \angle OBA = y$$. $$\angle AOB = 180 - 2y$$. We are given $$\angle ABC = 40^{\circ}$$.

It is most likely that the angle 40 degrees is $$\angle BAC$$. Then $$x = \angle BOC = 2 * \angle BAC = 2 * 40^{\circ} = 80^{\circ}$$.

Looking closely at the image, it seems that the angle 40 degrees is $$\angle OAB$$. If $$\angle OAB = 40^{\circ}$$, since OA=OB, then $$\angle OBA = 40^{\circ}$$. Then $$\angle AOB = 180 - (40+40) = 100^{\circ}$$.

If the angle 40 degrees is $$\angle OBA$$. Then since OA=OB, $$\angle OAB = 40^{\circ}$$. Then $$\angle AOB = 180 - (40+40) = 100^{\circ}$$.

Let's assume the angle 40 degrees is $$\angle BAC$$. Then $$\angle BOC = x = 2 * \angle BAC = 2 * 40^{\circ} = 80^{\circ}$$.

Given the placement of the angle marker, it is most likely $$\angle BAC = 40^{\circ}$$.

Alternate Segment Theorem: The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

Let's assume BC is a tangent at B. Then $$\angle ABC$$ is the angle between tangent BC and chord AB. The angle in the alternate segment is $$\angle ACB$$. So $$\angle ACB = 40^{\circ}$$.

In $$\triangle OBC$$, OB=OC (radii), so it is an isosceles triangle. $$\angle OBC = \angle OCB$$. We don't know $$\angle OBC$$.

Let's assume the angle 40 degrees is $$\angle OAB$$. Since OA = OB, $$\angle OBA = 40^{\circ}$$. Then $$\angle AOB = 180^{\circ} - (40^{\circ} + 40^{\circ}) = 100^{\circ}$$.

Let's assume the angle 40 degrees is $$\angle BAC$$. Then the angle subtended by arc BC at the center is $$\angle BOC = x$$. The angle subtended by arc BC at the circumference is $$\angle BAC$$. Thus, $$x = 2 imes \angle BAC = 2 imes 40^{\circ} = 80^{\circ}$$.

Given that the angle 40 degrees is placed near point A, it's highly probable that it represents $$\angle BAC$$.

In a circle, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Here, $$\angle BOC$$ is the angle at the center subtended by arc BC. $$\angle BAC$$ is the angle at the circumference subtended by arc BC.

Therefore, $$x = \angle BOC = 2 imes \angle BAC$$.

Assuming the angle marked $$40^{\circ}$$ is $$\angle BAC$$, then $$x = 2 imes 40^{\circ} = 80^{\circ}$$.

Final Answer: $$80^{\circ}$$