Illustration of different types of angles formed by intersecting rays from a central point.

Since no specific question was asked in the prompt, but the image is titled "Types of Angles" and shows rays OA, OB, OC, OD, OE, OF emanating from point O, I will assume the most common educational task associated with such a diagram: identify all pairs of perpendicular lines (or right angles) and possibly other angle types.

However, looking at the diagram:

- Ray OC is vertical upward.
- Ray OA is horizontal to the right.
- Ray OD is horizontal to the left.
- So, angle AOC = 90° → OA ⊥ OC
- Angle COD = 90° → OC ⊥ OD
- Angle DOA = 180° → straight line
- Ray OB is diagonal up-right — likely 45° from OA if symmetric.
- Ray OE is opposite to OB → so BOE is a straight line.
- Ray OF is diagonal down-right — likely 45° below OA.

Assuming standard positioning (as often used in textbooks):

- If OB bisects angle AOC, then ∠AOB = 45°, ∠BOC = 45°
- Similarly, if OF is symmetric to OB below OA, then ∠AOF = 45°
- Then OE is opposite OB → so ∠EOA = 135°, etc.

But without explicit measurements or labels, we must rely on visual cues.

In many such diagrams:

→ OC is perpendicular to AD (since AD is horizontal, OC vertical)
→ So angles between OC and OA, and OC and OD are right angles.

Also, if OB and OE form a straight line, and OF is symmetric to OB across OA, then:

∠BOF might be 90°? Let’s check:

If AOB = 45° and ∠AOF = 45°, then ∠BOF = 90° → so OB ⊥ OF

Similarly, if OE is opposite OB, then EOF = 135°, not 90°.

Wait — let’s assign coordinates for clarity (even though we won’t mention it):

Let O be origin.

- OA → positive x-axis
- OC → positive y-axis
- OD → negative x-axis
- OB → direction (1,1) → 45°
- OF → direction (1,-1) → -45° or 315°
- OE → direction (-1,-1) → 225°

Then:

- Vector OB • Vector OF = (1)(1) + (1)(-1) = 1 - 1 = 0 → so OB OF ✔
- Vector OC • Vector OA = (0,1)•(1,0)=0 → OC ⊥ OA ✔
- Vector OC • Vector OD = (0,1)•(-1,0)=0 → OC ⊥ OD ✔
- Vector OA • Vector OF = (1,0)•(1,-1)=1 ≠ 0 → not perpendicular
- Vector OB • Vector OE = (1,1)•(-1,-1) = -1 -1 = -2 ≠ 0 → not perpendicular

So perpendicular pairs are:

1. OA and OC
2. OC and OD
3. OB and OF

Also, since OA and OD are opposite, they form a straight line — not perpendicular.

Now, what about OE and OF?

Vector OE = (-1,-1), OF = (1,-1)

Dot product: (-1)(1) + (-1)(-1) = -1 + 1 = 0 → OE ⊥ OF ✔

Oh! So also OE ⊥ OF.

And OE and OB? Already checked — dot product = -2 → not perpendicular.

What about OC and OB? (0,1)•(1,1)=1 → not zero.

OC and OF? (0,1)•(1,-1)= -1 → not zero.

OA and OE? (1,0)•(-1,-1)= -1 → not zero.

OD and OB? (-1,0)•(1,1)= -1 → not zero.

OD and OF? (-1,0)•(1,-1)= -1 → not zero.

OD and OE? (-1,0)•(-1,-1)=1 → not zero.

So only these four perpendicular pairs:

- OA ⊥ OC
- OC ⊥ OD
- OB ⊥ OF
- OE ⊥ OF

Wait — OE ⊥ OF? Yes, as calculated.

But note: OF is shared in two perpendicular pairs: with OB and with OE.

That’s fine — one ray can be perpendicular to multiple others.

Also, is there any other? What about OC and... no.

Another way: look for right angles visually.

At point O:

- Between OA and OC: looks like 90°
- Between OC and OD: 90°
- Between OB and OF: if OB is 45° up, OF is 45° down, then angle between them is 90° → yes
- Between OE and OF: OE is 225°, OF is 315° → difference 90° → yes

So total 4 pairs of perpendicular rays.

But sometimes questions ask for “which lines are perpendicular” meaning full lines, not just rays.

Note that:

- Line AD (through OA and OD) is horizontal
- Line CF? Wait, C is up, F is down-right — not aligned.

Actually, ray OC is part of vertical line; ray OF is not vertical.

Better to think in terms of lines formed by opposite rays:

- Line AD: through A-O-D → horizontal
- Line CF? No, C and F are not opposite. Opposite of C would be... not drawn. But E is opposite B, F is not opposite anyone except maybe... wait, no ray opposite to C is shown. Only A-D, B-E, and C has no opposite, F has no opposite.

Actually, in the diagram:

- Rays: OA, OB, OC, OD, OE, OF
- Opposite pairs: OA & OD (same line), OB & OE (same line), but OC and OF have no opposites shown.

So lines present:

1. Line AD (horizontal)
2. Line BE (diagonal from bottom-left to top-right)
3. Ray OC (vertical up) — but no downward ray, so not a full line? In geometry problems, even if only one ray is drawn, if it's labeled as part of a line, we consider the infinite line.

But typically in such diagrams, when they draw ray OC upward, and no downward, it may imply the vertical line.

Similarly, OF is drawn downward-right, but no upward-left counterpart — unless OE is considered, but OE is opposite OB.

To avoid confusion, perhaps the intended answer is based on visible right angles.

Commonly in such figures, the expected answers are:

- ∠AOC = 90° → OA ⊥ OC
- ∠COD = 90° → OC ⊥ OD
- BOF = 90° → OB ⊥ OF
- And sometimes ∠EOF = 90°? From earlier calculation, yes.

But let me double-check ∠EOF:

Points: E is 225°, F is 315° → difference 90° → yes.

Similarly, ∠BOF: B is 45°, F is 315° → difference 270°, but smaller angle is min(270, 360-270)=90° → yes.

So both are valid.

Perhaps the problem expects listing all right angles or perpendicular pairs.

Since the user didn't specify the exact question, but given the context ("Types of Angles"), and the diagram, the most reasonable assumption is to identify all pairs of perpendicular lines/rays.

Final list of perpendicular pairs (rays forming right angles at O):

1. OA and OC
2. OC and OD
3. OB and OF
4. OE and OF

Is that all?

What about OC and... say, is there a ray perpendicular to OC besides OA and OD? OA and OD are both perpendicular to OC, but they are on the same line.

In terms of distinct lines:

- Line AD (horizontal) is perpendicular to line containing OC (vertical) → so one pair: AD ⊥ OC-line
- Line BE (from B to E) has slope 1 (if B is (1,1), E is (-1,-1))
- Line OF: from O to F (1,-1) → slope -1
- Slope of BE is 1, slope of OF is -1 → product = -1 → perpendicular → so line BE ⊥ line OF
- Also, line OF and line OE? OE is same as BE? No, OE is part of BE.

Line BE contains points B, O, E.

Ray OF is separate.

So line BE and ray OF: since OF is not on BE, and their directions are perpendicular, so the lines are perpendicular.

Similarly, line AD and line OC are perpendicular.

Additionally, is line OF perpendicular to anything else? We have OE ⊥ OF, but OE is on line BE, so already covered.

What about OC and OF? Not perpendicular.

So essentially, three pairs of perpendicular *lines*:

1. Line AD and line OC (assuming OC extends downward, even if not drawn)
2. Line BE and line OF
3. Also, since OE is on BE, and OE ⊥ OF, same as above.

But line OF is only a ray — does it define a line? Typically, we consider the infinite line through O and F.

Similarly for others.

To simplify for a student, perhaps list the ray pairs that form right angles.

Given that, and to match common textbook answers, I'll go with:

Perpendicular ray pairs (forming 90° angles at O):

- OA and OC
- OC and OD
- OB and OF
- OE and OF

Yes.

Some might argue that OA and OD are not both needed since they're collinear, but the angles are different: ∠AOC and ∠COD are both 90°, so both pairs are valid.

Now, for the final answer, since no specific question was asked, but the image is about "Types of Angles", and the most prominent feature is the right angles, I'll provide the list of perpendicular pairs.

If the actual homework question was something else (like "find measure of angle BOF"), but since it's not specified, I have to infer.

Looking back at the user input: "You uploaded an image containing a task." — but no text of the task. So I must assume the task is implied by the image title and content.

In many cases, such images are used to ask: "Identify all pairs of perpendicular lines" or "Name all right angles".

I think "name all right angles" is safer.

Right angles at O:

- ∠AOC
- ∠COD
- ∠BOF
- ∠EOF

Is ∠BOF the same as ∠FOB? Yes, angle naming doesn't matter order for magnitude.

Also, is there ∠COB? No, that's 45° if OB is at 45°.

Similarly, ∠AOF is 45°.

So only those four right angles.

But ∠EOF: points E, O, F — yes, 90°.

Similarly, ∠BOF: B, O, F — 90°.

Now, what about reflex angles? No, usually we take the smaller angle.

So final decision: the right angles are ∠AOC, ∠COD, ∠BOF, and ∠EOF.

To write it neatly.

Perhaps the problem is to find how many right angles, or list them.

I think listing them is best.

So for Final Answer, I'll put the list.

But let me confirm with a different approach.

Suppose we measure:

Assume OA is 0°, OC is 90°, OD is 180°, OB is 45°, OF is 315° (or -45°), OE is 225°.

Then angles between adjacent rays:

But better to compute all possible angles between pairs.

The angle between two rays is the absolute difference in their directions, taken as the smallest angle (≤180°).

So:

- OA (0°) and OC (90°): |90-0|=90° → right angle
- OA and OD (180°): 180° → straight
- OA and OB (45°): 45°
- OA and OF (315°): min(|315-0|, 360-315)=min(315,45)=45°
- OA and OE (225°): min(225, 135)=135°? |225-0|=225, 360-225=135 → so 135°
- OC (90°) and OD (180°): 90° → right angle
- OC and OB (45°): |90-45|=45°
- OC and OF (315°): |90-315|=225, min(225,135)=135°
- OC and OE (225°): |90-225|=135°
- OD (180°) and OB (45°): |180-45|=135°
- OD and OF (315°): |180-315|=135°
- OD and OE (225°): |180-225|=45°
- OB (45°) and OF (315°): |45-315|=270, min(270,90)=90° → right angle
- OB and OE (225°): |45-225|=180° → straight
- OB and OC: already did, 45°
- OF and OE (225°): |315-225|=90° → right angle
- OF and OC: 135°
- OE and OC: 135°
- OE and OD: 45°

So the right angles (90°) are between:

- OA and OC
- OC and OD
- OB and OF
- OE and OF

Exactly as before.

No others.

So four right angles.

Therefore, if the task is to identify all right angles, these are them.

If the task is to name perpendicular lines, then the lines containing these rays.

But for simplicity, since the diagram uses rays, I'll stick with ray pairs.

Final Answer should be the list.

Perhaps format as: The right angles are ∠AOC, ∠COD, ∠BOF, and ∠EOF.

To make it clear.

Since the user might expect a specific format, but none given, I'll go with this.

So after careful step-by-step reasoning, here is the final answer.