https://file.aiquickdraw.com/mathbot/file/1726398858310mazbx52v.jpg

To solve the problem in the image, we can break it down step by step.

### Part (i): Work out the size of the angle marked \(x\).
We are given that:
- \( \angle AOB = x \),
- \( \angle BOC = 90^\circ \),
- \( \angle AOC = 220^\circ \).

Since all three angles, \( \angle AOB \), \( \angle BOC \), and \( \angle AOC \), lie around the point \(O\) on a straight line (since \( OA \), \( OB \), and \( OC \) are straight lines), they should add up to \(360^\circ\). This is because angles around a point always sum to \(360^\circ\).

Thus, we have the equation:
\[
x + 90^\circ + 220^\circ = 360^\circ
\]
Simplifying:
\[
x + 310^\circ = 360^\circ
\]
Solving for \(x\):
\[
x = 360^\circ - 310^\circ = 50^\circ
\]

Therefore, \(x = 50^\circ\).

### Part (ii): Give a reason for your answer.
The reason is that the angles around a point must sum to \(360^\circ\). Since \( \angle AOB \), \( \angle BOC \), and \( \angle AOC \) are angles around the point \(O\), their sum must equal \(360^\circ\). By subtracting the sum of the other two angles (\(90^\circ\) and \(220^\circ\)) from \(360^\circ\), we can find the value of \(x\).

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Do you have any questions or need further clarification? Here are some related questions:

1. What is the rule for the sum of angles on a straight line?
2. How would the problem change if \( \angle BOC \) were \(100^\circ\) instead of \(90^\circ\)?
3. Can the sum of angles around a point ever be less than \(360^\circ\)?
4. What is the relationship between complementary and supplementary angles?
5. How would the calculation change if \( \angle AOC \) were \(180^\circ\) instead of \(220^\circ\)?

**Tip:** Always remember that the sum of angles around a point is \(360^\circ\), which is a crucial rule for solving angle problems.

OA, OB and OC are three straight lines. Work out the size of the angle marked x.

In this geometry problem, solve for the angle x using the sum of angles around a point, which must equal 360°. Given angles AOC = 220° and BOC = 90°, this problem is ideal for students in Grades 7-9.

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