Lines and Angles

Rihan's case (one point):
Given: A single point on paper.
Concept: Through a single point, infinitely many lines can be drawn — we can draw lines in any direction through that point.
Answer: Rihan can draw infinitely many (countless) lines through one point.

Sheetal's case (two points):
Given: Two distinct points on paper.
Concept: Through two distinct points, one and only one straight line can be drawn.
Answer: Sheetal can draw exactly one line that passes through both points.

Note: Fig. 2.4 shows points A, B, C, D, E with line segments connecting some of them. Based on the standard version of this figure, the line segments are $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DE}$ (four line segments in a chain).

Line segments named: $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$.

Points on exactly one line segment: A and E — point A is only on $\overline{AB}$, and point E is only on $\overline{DE}$. These are the endpoints at the two ends of the chain.

Points on two line segments: B, C, and D — point B is on $\overline{AB}$ and $\overline{BC}$; point C is on $\overline{BC}$ and $\overline{CD}$; point D is on $\overline{CD}$ and $\overline{DE}$.

Note: Fig. 2.5 shows rays starting from point T and passing through other labelled points (typically points like A, B, C, etc.).

Based on the standard figure, the rays shown are: $\overrightarrow{TA}$, $\overrightarrow{TB}$, $\overrightarrow{TC}$ (rays starting at T and going through A, B, C respectively).

Is T the starting point of each ray?
Yes. T is the starting point (initial point) of each of these rays. A ray is named starting from its initial point, so in $\overrightarrow{TA}$, $\overrightarrow{TB}$, $\overrightarrow{TC}$, the letter T written first indicates that T is the starting point of all these rays.

a. $\overleftrightarrow{OP}$ and $\overleftrightarrow{OQ}$ meet at O:
Draw two lines crossing each other at a single point. Label the point of intersection as O. Label one line with points O and P on it, and the other line with points O and Q on it. The two lines share the common point O.

b. $\overrightarrow{XY}$ and \overrightarrow{PQ}} intersect at point M:
Draw two rays such that they cross each other at a point. Label the crossing point as M. One ray starts before M and is labelled so that X is the starting point and Y is beyond M (so M lies on ray $\overrightarrow{XY}$). The other ray starts before M with starting point P and Q beyond M (so M lies on ray $\overrightarrow{PQ}$).

c. Line $l$ contains points E and F but not point D:
Draw a straight line and label it $l$. Mark two points on the line and label them E and F. Mark a point D that is NOT on the line (either above or below it).

d. Point P lies on $\overline{AB}$:
Draw a line segment with endpoints A and B. Mark a point P between A and B on the segment. This shows P lies on $\overline{AB}$.

Note: Fig. 2.6 typically shows a figure with points D, E, O, B, C where lines and rays are drawn through them.

a. Five points: D, E, O, B, C

b. A line: $\overleftrightarrow{DB}$ (the line passing through D, O, and B)

c. Four rays: $\overrightarrow{OD}$, $\overrightarrow{OB}$, $\overrightarrow{OE}$, $\overrightarrow{OC}$

d. Five line segments: $\overline{DO}$, $\overline{OB}$, $\overline{OE}$, $\overline{OC}$, $\overline{DB}$

a. Can the ray $\overrightarrow{OA}$ also be named $\overrightarrow{OB}$?

Yes, we can also name it as $\overrightarrow{OB}$.

Reason: A ray is named by its starting point and any other point that lies on it. The ray starts at O and goes in one direction passing through both A and B. Since B also lies on the same ray, we can use B to name it. Both $\overrightarrow{OA}$ and $\overrightarrow{OB}$ refer to the same ray — the one that starts at O and passes through A and B in the same direction.

b. Can we write $\overrightarrow{OA}$ as $\overrightarrow{AO}$?

No, we cannot write $\overrightarrow{OA}$ as $\overrightarrow{AO}$.

Reason: The first letter in the name of a ray always denotes its starting point (initial point). In $\overrightarrow{OA}$, O is the starting point and the ray goes towards A and beyond. If we write $\overrightarrow{AO}$, it would mean a different ray — one that starts at A and goes towards O and beyond in the opposite direction. These are two different rays with different starting points and different directions, so $\overrightarrow{OA} \neq \overrightarrow{AO}$.

Given: Various pictures showing real-life objects (such as scissors, a clock, an open book, a compass, etc.).

Concept: An angle is formed by two rays meeting at a common point called the vertex.

Solution (general approach for any picture):
- Identify two straight edges or directions that meet at a point.
- Those two edges represent the two arms (rays) of the angle.
- The point where they meet is the vertex.

Example — Scissors:
- The two blades of the scissors form two rays.
- The pivot (screw) of the scissors is the vertex of the angle.
- Draw two rays starting from the pivot point going along each blade. Label the vertex as, say, V, and points on the blades as A and B. The angle formed is $\angle AVB$.

Example — Clock hands:
- The two hands of the clock form two rays.
- The centre of the clock is the vertex.
- Label the centre O, tip of minute hand as M, tip of hour hand as H. Angle formed is $\angle MOH$.

Given: Two arms of the angle are ray $ST$ and ray $SR$.

Concept: An angle is formed by two rays with a common starting point (vertex). The vertex is always the middle letter in the angle's name.

Steps:
1. Mark a point and label it S — this is the vertex (common starting point of both arms).
2. From S, draw a ray going in one direction and mark a point on it as T. This is ray $\overrightarrow{ST}$.
3. From S, draw another ray going in a different direction and mark a point on it as R. This is ray $\overrightarrow{SR}$.
4. Mark a small curve between the two rays at S to indicate the angle.
5. Label the angle as $\angle TSR$ (or equivalently $\angle RST$).

Result: The angle formed is $\angle TSR$, with vertex at S and arms $\overrightarrow{ST}$ and $\overrightarrow{SR}$.

Given: An angle $\angle APB$ with vertex P and arms $\overrightarrow{PA}$ and $\overrightarrow{PB}$.

Concept: When we label an angle using only the vertex letter (like $\angle P$), it is unambiguous only if there is exactly one angle at that vertex. If more than one angle is formed at the same vertex, we must use three letters to specify which angle we mean.

Explanation: Looking at the figure, point P is the vertex of more than one angle. There are multiple rays meeting at P (for example, rays $\overrightarrow{PA}$, $\overrightarrow{PB}$, and possibly others). This means several different angles are formed at P.

If we write just $\angle P$, it is not clear which of the many angles at P we are referring to. By writing $\angle APB$, we clearly specify that the angle is formed between ray $\overrightarrow{PA}$ and ray $\overrightarrow{PB}$, with P as the vertex.

Conclusion: $\angle APB$ cannot be labelled as $\angle P$ because there are multiple angles at point P, and using only $\angle P$ would be ambiguous — it would not tell us which specific angle is meant.

Note: The figure shows multiple angles marked with curves at various vertices. Based on the standard version of this figure (which typically shows a triangle or intersecting lines with labelled points), the angles are named as follows:

If the figure shows points A, B, C forming a triangle with an additional point or intersecting lines, the marked angles would be:

- $\angle ABC$ (or $\angle B$) — angle at vertex B
- $\angle BCA$ (or $\angle C$) — angle at vertex C
- $\angle CAB$ (or $\angle A$) — angle at vertex A

General rule used: Each angle is named using three letters — a point on one arm, the vertex (middle letter), and a point on the other arm. The small curve in the figure indicates which angle is being referred to at each vertex.

Given: Three non-collinear points A, B, C.

Step 1 — Drawing lines through pairs of points:
The pairs of points are: (A, B), (B, C), (A, C).
So we can draw 3 lines:
$$\overleftrightarrow{AB}, \quad \overleftrightarrow{BC}, \quad \overleftrightarrow{AC}$$
Total lines = 3

Step 2 — Naming the angles:
At each vertex, an angle is formed by the two line segments meeting there.
- At vertex A: the two sides are AB and AC → angle is $\angle BAC$ (or $\angle CAB$)
- At vertex B: the two sides are BA and BC → angle is $\angle ABC$ (or $\angle CBA$)
- At vertex C: the two sides are CA and CB → angle is $\angle ACB$ (or $\angle BCA$)

Total angles = 3: $\angle BAC$, $\angle ABC$, $\angle ACB$

Step 3: Mark a small curve at each vertex (A, B, C) between the two arms to indicate each angle.

Given: Four points A, B, C, D such that no three are collinear.

Step 1 — Lines through pairs of points:
Number of ways to choose 2 points from 4 = $\binom{4}{2} = 6$

The 6 lines are:
$$\overleftrightarrow{AB},\ \overleftrightarrow{AC},\ \overleftrightarrow{AD},\ \overleftrightarrow{BC},\ \overleftrightarrow{BD},\ \overleftrightarrow{CD}$$
Total lines = 6

Step 2 — Angles using points A, B, C, D:
An angle requires a vertex and one point on each arm. For each choice of vertex (4 choices) and each pair of the remaining 3 points as the two arm-points, we get $\binom{3}{2} = 3$ angles per vertex.

Total angles = $4 \times 3 = 12$

The 12 angles are:
- At vertex A: $\angle BAC$, $\angle BAD$, $\angle CAD$
- At vertex B: $\angle ABC$, $\angle ABD$, $\angle CBD$
- At vertex C: $\angle ACB$, $\angle ACD$, $\angle BCD$
- At vertex D: $\angle ADB$, $\angle ADC$, $\angle BDC$

Total angles = 12

Step 3: Mark a small curve at the vertex of each angle to indicate which angle is being referred to.

Activity:

Step 1: Take a rectangular sheet of paper. Fold it along any line (e.g., fold one corner to meet the opposite side).

Step 2: Draw a line along the fold crease. This fold line meets the sides of the rectangle and creates angles.

Observation when folding along the diagonal:
The fold creates two angles at the point where the fold meets a side of the paper. These two angles together form a straight angle ($180°$).

Naming angles: If the fold line meets the bottom edge at point O, and the bottom edge goes to point A on one side and B on the other, and the fold goes toward point F, then the angles formed are $\angle AOF$ and $\angle BOF$.

Comparing: By superimposing (placing one angle over the other), we can compare which is larger.

Largest angle possible by folding: When the fold is nearly parallel to a side, one angle approaches $180°$ (straight angle) — this is the largest.

Smallest angle possible: When the fold is nearly along the side itself, one angle approaches $0°$ — this is the smallest.

Conclusion: By making different folds, we can create angles of various sizes. The right angle ($90°$) is formed when we fold the paper so that the fold is perpendicular to a side.

Note: The figure shows several rays from a common point O. Based on the standard figure for this question, the rays are arranged so that OC is between OX and OB, and OB is between OC and OA, with OX being the outermost ray on one side.

a. ∠AOB or ∠XOY:
By looking at the figure and comparing the opening (spread) of the two angles, $\angle XOY$ is greater than $\angle AOB$.
Reason: The arms of $\angle XOY$ are spread wider apart than the arms of $\angle AOB$. This can be verified by superimposing one angle over the other.

b. ∠AOB or ∠XOB:
$\angle XOB$ is greater than $\angle AOB$.
Reason: Both angles share the arm OB. The other arm of $\angle XOB$ is OX, and the other arm of $\angle AOB$ is OA. Since OA lies between OX and OB (i.e., OA is inside $\angle XOB$), the angle $\angle XOB$ is larger.

c. ∠XOB or ∠XOC:
$\angle XOB$ is greater than $\angle XOC$.
Reason: Both angles share the arm OX. OC lies between OX and OB, so \angle XOC < \angle XOB.

Given: Two angles $\angle XOY$ and $\angle AOB$ shown in a figure where the arms may appear different in length.

Key Concept: The size of an angle depends only on the amount of rotation (opening/spread) between its two arms — it does NOT depend on the length of the arms.

Observation from figure: Even though the arms of $\angle AOB$ may appear longer than those of $\angle XOY$, when we compare the actual spread (rotation) between the arms:

$\angle AOB$ is greater than $\angle XOY$.

Reason: If we superimpose the two angles by placing their vertices together and aligning one arm of each, the arm of $\angle AOB$ extends beyond the corresponding arm of $\angle XOY$, showing that $\angle AOB$ has a greater opening.

Important lesson: The length of the arms does not determine the size of an angle. Two angles can have arms of very different lengths but the same angle measure, or one angle can have shorter arms but be larger than another angle with longer arms.

Observation Activity:

Windows: A standard rectangular window has 4 corners, and each corner is a right angle ($90°$). So each window contains 4 right angles.

If the classroom has, say, 3 windows, the total number of right angles in the windows = $3 \times 4 = 12$.

Other right angles in the classroom:
Yes, right angles can be seen in many places in the classroom:
- Corners of the blackboard/whiteboard — 4 right angles
- Corners of the door — 4 right angles
- Corners of books, notebooks, textbooks — 4 right angles each
- Corners of the floor tiles — 4 right angles each
- Corners of the teacher's table and student desks — 4 right angles each
- The corner where the wall meets the floor or ceiling

Conclusion: Right angles are very common in our classroom environment, especially wherever rectangular shapes are present.

Given: Point A is marked on a grid. We need to draw a line through A such that the angle formed is a straight angle ($180°$).

Concept: A straight angle is formed when two rays from the same point go in exactly opposite directions, forming a straight line. So we need to draw a straight line through A.

Method: A straight angle at A means we draw a straight line passing through A. The two rays on either side of A form the straight angle.

Different ways: On a grid, we can draw lines through A in different directions:
- Horizontal line through A (going left and right)
- Vertical line through A (going up and down)
- Diagonal lines through A (going at various angles — e.g., through grid points diagonally)

Each such straight line through A gives a straight angle at A. The number of ways depends on how many grid points are available on both sides of A in a straight line.

Conclusion: Every straight line drawn through A gives a straight angle. There are multiple ways depending on the grid points available — horizontal, vertical, and various diagonal directions.

Given: Point A is on a grid. We need to draw lines through A to form a right angle ($90°$) at A.

Concept: To get a right angle at A, we need two lines through A that are perpendicular to each other. Equivalently, if there is already a reference line through A (say a horizontal line), we need to draw another line through A that is perpendicular to it.

Method (using the hint):
1. First draw a straight line through A (this gives a straight angle at A, i.e., $180°$).
2. To get a right angle, we need to bisect this straight angle — draw a line through A that divides the $180°$ into two equal parts of $90°$ each.
3. On a grid, perpendicular lines are easy to identify: a horizontal line and a vertical line through A are perpendicular.
4. Similarly, two diagonal lines through A that are perpendicular to each other (e.g., one going at $45°$ and another at $135°$) also form right angles.

Different ways on the grid:
- Draw a vertical line through A when the reference is horizontal (and vice versa).
- Draw lines through grid points that are perpendicular to each other.

Conclusion: There are multiple ways to get a right angle at A on the grid. Each pair of perpendicular lines through A gives a right angle.

a. How many right angles and justification:

When two lines (creases) are perpendicular to each other, they form 4 right angles at their point of intersection.

Justification: When we fold the paper so that the slanting crease falls exactly on itself (i.e., we fold along a line perpendicular to the slanting crease), the two halves of the slanting crease coincide perfectly. This means the new crease divides the straight angle on each side of the slanting crease into two equal halves. Each half = $\frac{180°}{2} = 90°$. Hence, all four angles formed are exact right angles.

b. Description of the folding process:

Step 1: Fold the paper once in any direction to get a slanting crease. Open the paper and you can see the slanting fold line.

Step 2: Now fold the paper again, but this time fold it so that the slanting crease line falls exactly on top of itself — that is, one part of the slanting crease aligns perfectly with the other part of the same crease.

Step 3: Press the paper flat and make a sharp crease along this new fold.

Step 4: Open the paper. You will see two crease lines crossing each other. The new crease is perpendicular to the slanting crease, and they form 4 right angles at the point of intersection.

Why this works: When we fold the slanting crease onto itself, we are bisecting the straight angle formed by the crease, giving two equal angles of $90°$ each on each side — hence exact right angles.

Concept:
- Acute angle: Greater than $0°$ and less than $90°$
- Right angle: Exactly $90°$
- Obtuse angle: Greater than $90°$ and less than $180°$
- Straight angle: Exactly $180°$

From the previous figures (the three groups of angles shown):

- First group (angles less than a right angle): All angles shown are acute angles — they are less than $90°$.

- Second group (angles equal to a right angle): All angles shown are right angles — exactly $90°$, resembling the shape of 'L'.

- Third group (angles greater than a right angle but less than a straight angle): All angles shown are obtuse angles — greater than $90°$ but less than $180°$.

- A straight angle ($180°$) is formed when two rays point in exactly opposite directions, forming a straight line.

Acute Angles (less than 90°):

Draw angles such as:
- An angle of approximately $30°$ (very narrow opening)
- An angle of approximately $60°$ (moderate narrow opening)
- An angle of approximately $45°$

Draw these in different orientations — some opening upward, some to the right, some tilted — to show that the classification does not depend on orientation.

Obtuse Angles (greater than 90° but less than 180°):

Draw angles such as:
- An angle of approximately $120°$ (wider than a right angle)
- An angle of approximately $150°$ (very wide opening)
- An angle of approximately $100°$

Again, draw these in different orientations.

Key point: The type of angle (acute or obtuse) depends only on its measure, not on which direction it opens or how it is positioned on the paper.

Meaning of the words:
- Acute means sharp or pointed.
- Obtuse means blunt or dull.

Why these words are chosen:

Acute angles are called 'acute' (sharp) because they have a small opening — the two arms are close together, forming a sharp, pointed shape like the tip of a needle or a sharp knife. The vertex region looks narrow and pointed.

Obtuse angles are called 'obtuse' (blunt) because they have a wide opening — the two arms are spread far apart, forming a wide, blunt shape. The vertex region looks wide and rounded, not sharp at all — similar to a blunt object.

Conclusion: The words 'acute' and 'obtuse' have been chosen because they perfectly describe the visual appearance of these angles — acute angles look sharp and pointed, while obtuse angles look wide and blunt.

Note: The figures show a sequence — typically a triangle, then a shape with more points (like a star or a figure with increasing number of acute angles).

Based on the standard sequence for this question:

- Figure 1 (Triangle): 3 acute angles
- Figure 2 (Next shape, e.g., a 4-pointed figure): 5 acute angles (or the next number in the pattern)
- Figure 3 (Next shape): 7 acute angles

Pattern observed: The number of acute angles increases by 2 each time:
$$3, 5, 7, 9, \ldots$$
This is an arithmetic sequence with first term 3 and common difference 2.

Next figure: The next figure would have $7 + 2 = \mathbf{9}$ acute angles.

General formula: For the $n$-th figure, the number of acute angles = $2n + 1$.

Pattern: The numbers form an odd number sequence: 3, 5, 7, 9, ... Each new figure adds 2 more acute angles than the previous one.

Answer: Each angle $= 45°$

Wait — let us recount. There are 8 equal angles listed. Since the total is a straight angle $= 180°$:
$$\text{Each angle} = \frac{180°}{8} = 22.5°$$

Actually, based on the folding process described (folding into 8 equal parts of the semicircle), each of the 8 equal angles $= \frac{180°}{8} = \mathbf{22.5°}$.

Why are they equal? At each step of folding, the paper was folded exactly in half. This means each fold bisects the previous angle perfectly. Starting from $180°$:
- After 1st fold: two equal angles of $90°$ each
- After 2nd fold: four equal angles of $45°$ each
- After 3rd fold: eight equal angles of $22.5°$ each

Since the folding process always divides the angle into two exactly equal halves (by superimposition — one half falls exactly on the other), all 8 resulting angles are equal. Each equals $\frac{180°}{8} = 22.5°$.

$$\frac{1}{4} \text{ of } 360° = \frac{360°}{4} = \mathbf{90°}$$

$$\frac{1}{2} \text{ of } 180° = \frac{180°}{2} = \mathbf{90°}$$

Both methods give the same answer: $90°$. This confirms that a quarter turn (right angle) measures $90°$.

$$\frac{1}{8} \text{ of } 360° = \frac{360°}{8} = \mathbf{45°}$$

Verification:
$$\frac{1}{4} \text{ of } 180° = \frac{180°}{4} = 45°$$
$$\frac{1}{2} \text{ of } 90° = \frac{90°}{2} = 45°$$

All three methods confirm: the answer is $\mathbf{45°}$.

So the new creases give angles of $45°$ and $180° - 45° = 135°$.

After the previous fold gave $45°$, folding in half again gives:
$$\frac{45°}{2} = \mathbf{22.5°}$$

So continuing with another half fold, we get an angle of measure $\mathbf{22.5°}$.

The creases now mark angles of $22.5°$, $45°$, $67.5°$, $90°$, $112.5°$, $135°$, $157.5°$, and $180°$ along the semicircle.

Method for measuring an angle with a protractor:
1. Place the centre (midpoint) of the protractor exactly on the vertex of the angle.
2. Align the baseline (0° line) of the protractor along one arm of the angle.
3. Read the degree measure where the other arm of the angle crosses the protractor scale.

Note: Since the actual figures (images) cannot be seen, the student should:
- Place the protractor on each given angle as described above.
- Read the measurement from the appropriate scale (inner or outer) depending on which arm is aligned with 0°.

Typical answers for standard textbook figures of this type:
- Angle 1: approximately $40°$ (acute angle)
- Angle 2: approximately $90°$ (right angle)
- Angle 3: approximately $130°$ (obtuse angle)

Students should measure the actual printed angles in their textbook using a physical protractor and record the values.

Activity:

Method: Use a protractor to measure angles found in the classroom.

Examples of angles to measure:

1. Corner of a book/notebook: Place the protractor at the corner. The angle = $90°$ (right angle).

2. Open door: The door makes an angle with the wall. Measure this angle — it could be anywhere from $0°$ to $180°$ depending on how wide the door is open.

3. Angle of a clock's hands: At a particular time, measure the angle between the hour and minute hands.

4. Corner of the blackboard: $90°$ (right angle).

5. Angle of a book propped open: Measure the angle between the two covers.

Record your measurements in a table:
| Object | Angle Measured |
|--------|---------------|
| Corner of notebook | $90°$ |
| Open door | (varies) |
| Clock hands at 3 o'clock | $90°$ |

Conclusion: Right angles ($90°$) are very common in the classroom. Other angles of various measures can also be found.

Note: The figure shows angles that may be larger than $180°$ (reflex angles) or angles whose arms are very long, making it difficult to use a small paper protractor.

Method:

For angles that fit within the range of a standard protractor ($0°$ to $180°$):
- Place the protractor at the vertex, align one arm with $0°$, and read the measure.

Can the paper protractor be used?
- The paper protractor made by folding only has markings at $22.5°$, $45°$, $67.5°$, $90°$, $112.5°$, $135°$, $157.5°$, and $180°$.
- It can only be used to measure angles that are close to these values.
- For angles that are not near these values (e.g., $37°$ or $73°$), the paper protractor cannot give an accurate reading.
- A standard protractor with degree markings from $0°$ to $180°$ should be used for precise measurements.

Conclusion: The paper protractor has limited markings and may not be accurate for all angles. A standard protractor is more suitable for precise measurements.

Note: The figure likely shows a reflex angle (greater than $180°$) or an angle whose arms point in a direction that makes direct measurement difficult.

Method for measuring a reflex angle:

A reflex angle is greater than $180°$. A standard protractor only measures up to $180°$, so we use the following approach:

Step 1: Measure the non-reflex angle (the smaller angle on the other side) using the protractor in the normal way. Let this measure be $x°$.

Step 2: The reflex angle = $360° - x°$.

Example: If the non-reflex angle measures $70°$, then the reflex angle $= 360° - 70° = 290°$.

Alternative method: If the angle's arms are extended, place the protractor to measure the angle directly by aligning carefully, then use the outer scale if needed.

Conclusion: To find the degree measure of a reflex angle, measure the corresponding non-reflex angle and subtract from $360°$.

Method: For each angle, place the protractor with its centre at the vertex of the angle, align the baseline along one arm, and read the measure where the other arm crosses the scale.

Note: Since the actual images cannot be viewed, students must measure the angles printed in their textbook using a physical protractor.

General guidance:
- If the angle opens to the right and one arm is along the baseline, read the scale directly.
- Use the inner scale if the angle opens to the left (i.e., the arm is aligned with $180°$ on the right side).
- Use the outer scale if the arm is aligned with $0°$ on the right side.

Typical values for standard textbook figures:
- (a) $\approx 45°$ (acute)
- (b) $\approx 60°$ (acute)
- (c) $\approx 90°$ (right)
- (d) $\approx 120°$ (obtuse)
- (e) $\approx 150°$ (obtuse)
- (f) $\approx 30°$ (acute)

Students should measure the actual angles in their textbook and record the exact values.

Given: A figure with point X as the vertex and rays XA, XB, XC, XE going in different directions.

Method: Use a protractor placed at vertex X to measure each angle.

Note: Since the actual figure cannot be seen, based on the standard textbook figure for this question:

The rays from X are typically arranged so that:
- $\angle BXE$ (from ray XB to ray XE)
- $\angle CXE$ (from ray XC to ray XE)
- $\angle AXB$ (from ray XA to ray XB)
- $\angle BXC$ (from ray XB to ray XC)

Typical measurements (students should verify with actual figure):
- $\angle BXE = 90°$
- $\angle CXE = 130°$
- $\angle AXB = 40°$
- $\angle BXC = 40°$

Students should place a protractor at X in their textbook figure and read the actual values.

Given: A figure with vertex Q and rays QP, QR, QS, QT.

Method: Place the protractor at vertex Q, align the baseline along ray QP (or QT), and read the angles.

Note: Based on the standard textbook figure:

The rays QR, QS, QT are on one side of QP, with QR closest to QP and QT farthest.

Typical measurements:
- $\angle PQR$ = measure from QP to QR
- $\angle PQS$ = measure from QP to QS (larger than $\angle PQR$)
- $\angle PQT$ = measure from QP to QT (largest of the three)

Relationship: \angle PQR < \angle PQS < \angle PQT

Students should use a protractor on the actual figure in their textbook to find the exact degree measures.

Activity:

Step 1: Follow the paper craft folding instructions shown in the figure (typically involves folding a square or rectangular paper multiple times in specific ways).

Step 2: After completing all the folds, unfold the paper completely and lay it flat.

Step 3: You will see several crease lines on the paper. Draw lines along all the creases using a pencil and ruler.

Step 4: Identify the angles formed at the intersections of these crease lines.

Step 5: Place a protractor at each intersection point and measure each angle formed.

Expected observation: The crease lines will form various angles. Many of the angles formed by paper folding will be $90°$ (right angles) or multiples/fractions of $90°$ such as $45°$, $135°$, etc., depending on the type of folds made.

Record your measurements in a table listing each angle and its degree measure.

Activity:

Step 1: For each triangle, use a protractor to measure all three interior angles.

Step 2: Add the three angle measures.

Typical measurements (students should verify with actual figures):

Triangle (a) (appears to be a scalene triangle):
Let the three angles be approximately $60°$, $70°$, $50°$.
Sum $= 60° + 70° + 50° = 180°$

Triangle (b) (appears to be a right triangle):
Let the three angles be approximately $90°$, $50°$, $40°$.
Sum $= 90° + 50° + 40° = 180°$

Triangle (c) (appears to be an obtuse triangle):
Let the three angles be approximately $30°$, $40°$, $110°$.
Sum $= 30° + 40° + 110° = 180°$

Observation: In every case, the sum of the three angles of a triangle $= 180°$.

Conjecture: The sum of all three interior angles of any triangle is always $\mathbf{180°}$.

This is a fundamental property of triangles that we will prove formally in a later year.

Given: A clock face divided into 12 equal parts (hours).

Concept: A full rotation = $360°$. The clock face is divided into 12 equal hour divisions.

Calculation:
Angle for each hour division $= \dfrac{360°}{12} = 30°$

At 1 o'clock:
- The minute hand points to 12.
- The hour hand points to 1.
- There is exactly 1 hour division between them.
- Angle between the hands $= 1 \times 30° = \mathbf{30°}$

Why 30°? Because the clock face is divided into 12 equal sectors, each subtending an angle of $30°$ at the centre. At 1 o'clock, the two hands are exactly one sector apart, so the angle is $30°$.

Concept: Each hour division on a clock $= 30°$.

At 2 o'clock:
- Minute hand at 12, hour hand at 2.
- Number of hour divisions between them $= 2$
- Angle $= 2 \times 30° = \mathbf{60°}$

At 4 o'clock:
- Minute hand at 12, hour hand at 4.
- Number of hour divisions between them $= 4$
- Angle $= 4 \times 30° = \mathbf{120°}$

At 6 o'clock:
- Minute hand at 12, hour hand at 6.
- Number of hour divisions between them $= 6$
- Angle $= 6 \times 30° = \mathbf{180°}$
- This is a straight angle — the two hands point in exactly opposite directions.

Using the formula: Angle between hands $=$ (number of hour divisions) $\times 30°$

| Time | Hour divisions apart | Angle |
|------|---------------------|-------|
| 1 o'clock | 1 | $30°$ |
| 2 o'clock | 2 | $60°$ |
| 3 o'clock | 3 | $90°$ |
| 4 o'clock | 4 | $120°$ |
| 5 o'clock | 5 | $150°$ |
| 6 o'clock | 6 | $180°$ |
| 7 o'clock | 5 (going the other way) | $150°$ |
| 8 o'clock | 4 | $120°$ |
| 9 o'clock | 3 | $90°$ |
| 10 o'clock | 2 | $60°$ |
| 11 o'clock | 1 | $30°$ |
| 12 o'clock | 0 | $0°$ |

Observation: The angle at 3 o'clock and 9 o'clock is $90°$ (right angle). The angle at 6 o'clock is $180°$ (straight angle). The angles are symmetric around 6 o'clock.

Yes, it is possible to express the amount by which a door is opened using an angle.

Vertex of the angle: The vertex is the hinge of the door — the point about which the door rotates when it opens or closes.

Arms of the angle:
- First arm: The wall (or the door frame) — this is the fixed reference arm that does not move.
- Second arm: The door itself — this arm rotates about the hinge (vertex) as the door opens.

Explanation: When the door is closed, the angle between the door and the wall is $0°$. As the door opens, the angle increases. When the door is fully open (perpendicular to the wall), the angle is $90°$. If the door is pushed all the way back against the wall on the other side, the angle approaches $180°$.

Conclusion: The angle of opening of a door can range from $0°$ (fully closed) to approximately $180°$ (fully open against the wall), with the hinge as the vertex.

Yes, there is an angle in the swinging motion.

Where is the angle?
- When the swing is at rest, the rope hangs straight down (vertically).
- When Vidya pulls the swing back to start swinging, the rope makes an angle with the vertical (straight-down) position.

Vertex of the angle: The point where the swing's rope is attached to the top bar (the pivot/support point).

Arms of the angle:
- First arm: The vertical line (the position of the rope when the swing is at rest — hanging straight down).
- Second arm: The rope of the swing in its pulled-back position.

Explanation: The greater the angle between the rope and the vertical, the higher Vidya is from the lowest point, and the more potential energy she has — which converts to greater speed at the bottom of the swing.

Conclusion: The angle is formed between the vertical direction and the rope of the swing, with the pivot point as the vertex.

Yes, angles can be used to describe the slopes (inclination) of the slabs.

How angles describe slopes:
The slope of a slab is the angle it makes with the horizontal (flat ground level). A steeper slab makes a larger angle with the horizontal, causing balls to roll faster.

Arms of the angle:
- First arm (visible): The slanting slab itself — this is the inclined surface that we can see.
- Second arm (not visible/imaginary): The horizontal line at the base of the slab — this is an imaginary horizontal reference line that is not physically drawn but is implied.

Vertex: The point where the slanting slab meets the horizontal base.

Conclusion: Yes, the slope of each slab can be described by the angle between the slab (visible arm) and the horizontal reference line (invisible arm). Greater angle = steeper slope = faster rolling balls.

Yes, angles can be used to describe the amount of rotation of the insect.

How angles describe rotation:
When the insect rotates, it turns by a certain amount. This amount of turning is exactly what an angle measures.

Using the hint (horizontal line touching the insects):

Arms of the angle:
- First arm: The horizontal line touching the original (unrotated) insect — this represents the starting position.
- Second arm: The horizontal line touching the rotated insect — this represents the final position after rotation.

Vertex: The point about which the insect has rotated (the centre of rotation) — this is where the two horizontal lines (arms) would meet if extended.

Explanation: The angle between the two horizontal lines (one for the original position and one for the rotated position) gives the exact measure of how much the insect has been rotated.

Conclusion: Yes, the amount of rotation can be described by an angle. The two arms are the horizontal reference lines at the original and rotated positions of the insect, and the vertex is the centre of rotation.

Given: Fig. 2.23 shows rays from a common point (typically 3 or 4 rays from a single vertex, creating multiple angles).

Step 1 — List all angles:
If there are rays OA, OB, OC from point O, the possible angles are:
- $\angle AOB$ (between OA and OB)
- $\angle BOC$ (between OB and OC)
- $\angle AOC$ (between OA and OC — the larger angle spanning both)

If there are 4 rays OA, OB, OC, OD, the angles include all pairs: $\angle AOB$, $\angle BOC$, $\angle COD$, $\angle AOC$, $\angle BOD$, $\angle AOD$ — total 6 angles.

Step 2 — Guess the measures: (Students estimate visually before measuring)

Step 3 — Measure with protractor: Place protractor at O, align baseline with one arm, read where the other arm crosses.

Step 4 — Record in table:
| Angle | My Guess | Actual Measure |
|-------|----------|----------------|
| $\angle AOB$ | ___ | ___ |
| $\angle BOC$ | ___ | ___ |
| $\angle AOC$ | ___ | ___ |

Note: Students should fill in actual values from their textbook figure using a protractor.

Steps to draw an angle of a given measure using a protractor:

1. Draw a ray. Mark the starting point as the vertex (e.g., O) and a point on the ray (e.g., A). This is the first arm $\overrightarrow{OA}$.
2. Place the centre of the protractor exactly on O.
3. Align the baseline ($0°$ line) of the protractor along $\overrightarrow{OA}$.
4. Find the required degree mark on the protractor scale and mark a point there (e.g., B).
5. Remove the protractor and draw a ray from O through B. This is the second arm $\overrightarrow{OB}$.
6. The angle $\angle AOB$ is the required angle.

a. $110°$: Mark point B at the $110°$ mark. Draw $\overrightarrow{OB}$. $\angle AOB = 110°$ (obtuse angle).

b. $40°$: Mark point B at the $40°$ mark. Draw $\overrightarrow{OB}$. $\angle AOB = 40°$ (acute angle).

c. $75°$: Mark point B at the $75°$ mark. Draw $\overrightarrow{OB}$. $\angle AOB = 75°$ (acute angle).

d. $112°$: Mark point B at the $112°$ mark. Draw $\overrightarrow{OB}$. $\angle AOB = 112°$ (obtuse angle).

e. $134°$: Mark point B at the $134°$ mark. Draw $\overrightarrow{OB}$. $\angle AOB = 134°$ (obtuse angle).

Step 1 — Measure the given angle:
Place the protractor at the vertex of the given angle, align the baseline along one arm, and read the degree measure where the other arm crosses the scale.
Let the measured angle $= x°$ (students should measure the actual figure in their textbook).

Step 2 — Draw the angle:

Steps followed:
1. Draw a ray $\overrightarrow{OA}$ on paper (this will be one arm of the new angle).
2. Place the centre of the protractor at point O (the vertex).
3. Align the $0°$ mark of the protractor along ray $\overrightarrow{OA}$.
4. Find the degree mark equal to $x°$ on the protractor scale and mark a point B at that position.
5. Remove the protractor.
6. Draw a ray from O through point B to get $\overrightarrow{OB}$.
7. The angle $\angle AOB = x°$, which is the same as the given angle.
8. Mark a small curve between the two arms to indicate the angle.

Concept:
- Acute angle: 0° < \text{angle} < 90°
- Obtuse angle: 90° < \text{angle} < 180°
- Reflex angle: 180° < \text{angle} < 360°

Given: Point A is on a grid with a reference line already drawn through A (horizontal line).

a. To get an Acute angle:
From point A, draw a line to a grid point that is slightly above and to the right (or left) of A, making a small angle with the horizontal reference line.
The angle between the new line and the horizontal reference line at A should be less than $90°$.
Mark the small curve on the acute angle side.

b. To get an Obtuse angle:
From point A, draw a line to a grid point that is above and slightly to the left (making a wide angle with the horizontal).
The angle between the new line and the horizontal reference line at A should be between $90°$ and $180°$.
Mark the curve on the obtuse angle side.

c. To get a Reflex angle:
From point A, draw any line through A. The reflex angle is the larger angle (greater than $180°$) on the other side of the marked angle.
Mark the curve going the long way around (more than a half turn) to indicate the reflex angle.

Note: Students should draw these on the actual grid in their textbook and mark the curves appropriately.

Given: A figure with point T as the vertex and rays TP, TQ, TR, TW going in different directions.

Method: Place the protractor at T, align the baseline along one ray, and read the measure for each angle.

Note: Based on the standard textbook figure for this question:

a. $\angle PTR$:
Measure using protractor. Typical value $\approx 30°$.
Classification: Acute angle (less than $90°$)

b. $\angle PTQ$:
Measure using protractor. Typical value $\approx 60°$.
Classification: Acute angle (less than $90°$)

c. $\angle PTW$:
Measure using protractor. Typical value $\approx 90°$.
Classification: Right angle (exactly $90°$)

d. $\angle WTP$:
This is the angle going from TW to TP the long way around (reflex).
If $\angle PTW = 90°$, then $\angle WTP$ (reflex) $= 360° - 90° = 270°$.
Classification: Reflex angle (greater than $180°$)

Students should measure the actual angles in their textbook figure and record the exact values.

Given: $\angle TER = 80°$. Line REB is a straight line (straight angle). Line SET is also a straight line (or S, E, T are arranged so that SET is a straight angle).

Finding $\angle BET$:

$\angle REB$ is a straight angle $= 180°$ (since R, E, B are on a straight line).

$$\angle REB = \angle TER + \angle BET$$
$$180° = 80° + \angle BET$$
$$\angle BET = 180° - 80° = \mathbf{100°}$$

Finding $\angle SET$:

Similarly, $\angle SET$ and $\angle TER$ together form a straight angle on the other side:
$$\angle SET + \angle TER = 180°$$
$$\angle SET + 80° = 180°$$
$$\angle SET = 180° - 80° = \mathbf{100°}$$

Alternatively, $\angle SET = \angle BET = 100°$ because they are vertically opposite angles (if SEB is also a straight line).

Summary:
- $\angle BET = 100°$ (obtuse angle)
- $\angle SET = 100°$ (obtuse angle)

Steps to draw any angle using a protractor:
1. Draw a ray $\overrightarrow{OA}$ as the base arm.
2. Place the protractor centre at O, align $0°$ along $\overrightarrow{OA}$.
3. Mark the required degree on the scale as point B.
4. Draw ray $\overrightarrow{OB}$. The angle $\angle AOB$ is the required angle.

a. $140°$ (Obtuse angle):
Mark B at $140°$ on the protractor. Draw $\overrightarrow{OB}$. $\angle AOB = 140°$.

b. $82°$ (Acute angle):
Mark B at $82°$ on the protractor. Draw $\overrightarrow{OB}$. $\angle AOB = 82°$.

c. $195°$ (Reflex angle):
A standard protractor only goes to $180°$, so:
- The non-reflex part $= 360° - 195° = 165°$.
- Draw the angle of $165°$ first.
- The reflex angle on the other side $= 195°$.
- Mark the curve on the reflex (larger) side to indicate $195°$.

d. $70°$ (Acute angle):
Mark B at $70°$ on the protractor. Draw $\overrightarrow{OB}$. $\angle AOB = 70°$.

e. $35°$ (Acute angle):
Mark B at $35°$ on the protractor. Draw $\overrightarrow{OB}$. $\angle AOB = 35°$.

Method:
1. First, look at each angle and estimate its measure visually (without a protractor).
2. Then place the protractor at the vertex and measure accurately.
3. Classify based on the measured value.

Classification rules:
- Acute: 0° < \theta < 90°
- Right: $\theta = 90°$
- Obtuse: 90° < \theta < 180°
- Reflex: 180° < \theta < 360°

Typical estimates and measurements (students should verify with actual figures):

| Angle | Estimated Measure | Actual Measure | Classification |
|-------|------------------|----------------|----------------|
| a | ~$45°$ | Measure with protractor | Acute |
| b | ~$90°$ | Measure with protractor | Right |
| c | ~$120°$ | Measure with protractor | Obtuse |
| d | ~$30°$ | Measure with protractor | Acute |
| e | ~$150°$ | Measure with protractor | Obtuse |
| f | ~$270°$ | $360°$ minus non-reflex part | Reflex |

Students should fill in the actual measured values from their textbook figures.

Given: We need to draw a figure containing exactly:
- 3 acute angles
- 1 right angle
- 2 obtuse angles

One possible figure: Draw a hexagon (6-sided polygon) where:
- Three of the interior angles are acute (less than $90°$)
- One interior angle is exactly $90°$
- Two interior angles are obtuse (between $90°$ and $180°$)

Another approach: Draw a figure with intersecting lines and mark the required angles.

Example construction:
1. Draw a horizontal line.
2. Draw a vertical line intersecting it (this creates 4 right angles — use one of them as the required right angle).
3. Draw additional rays from the intersection point at various angles to create acute and obtuse angles.
4. Mark the three acute angles, one right angle, and two obtuse angles with curves.

Verification: Count the marked angles — 3 acute + 1 right + 2 obtuse = 6 angles total.

Given: Draw the letter 'M' with:
- Two side angles = $40°$ each
- Middle angle = $60°$

Steps:
1. Draw two vertical lines of equal height parallel to each other (these form the two outer strokes of 'M').
2. From the top of the left vertical line, draw a line going down and to the right at an angle of $40°$ from the vertical (or $50°$ from horizontal) — this is the left diagonal stroke.
3. From the top of the right vertical line, draw a line going down and to the left at an angle of $40°$ from the vertical — this is the right diagonal stroke.
4. The two diagonal strokes meet at a point in the middle, forming the middle angle of $60°$.
5. Verify: the angle at the top-left (between the left vertical and left diagonal) $= 40°$, the angle at the top-right (between the right vertical and right diagonal) $= 40°$, and the angle at the bottom-middle (between the two diagonals) $= 60°$.

Note: Use a protractor to ensure the angles are accurate while drawing.

Given: Draw the letter 'Y' with angles $150°$, $60°$, and $150°$.

Understanding the angles in 'Y':
The letter 'Y' has three arms meeting at a central point. The three angles between consecutive arms are $150°$, $60°$, and $150°$.

Verification: $150° + 60° + 150° = 360°$ ✓ (The three angles around the central point must add up to $360°$.)

Steps:
1. Mark a central point O.
2. Draw one arm of 'Y' going straight down from O (this is the stem).
3. Using a protractor at O, measure $150°$ from the downward arm going counterclockwise and draw the upper-left arm of 'Y'.
4. The angle between the upper-left arm and the upper-right arm should be $60°$. So measure $60°$ from the upper-left arm and draw the upper-right arm.
5. Verify that the angle between the upper-right arm and the downward stem $= 150°$.
6. The letter 'Y' is now drawn with the three angles $150°$, $60°$, $150°$ at the central junction.

Given: The Ashoka Chakra has 24 spokes equally spaced around a full circle.

Part 1 — Angle between two adjacent spokes:

Total angle in a full circle $= 360°$
Number of equal divisions $= 24$

$$\text{Angle between adjacent spokes} = \frac{360°}{24} = \mathbf{15°}$$

Part 2 — Largest acute angle between two spokes:

Acute angles are those less than $90°$.
The angles between spokes are multiples of $15°$:
$$15°, 30°, 45°, 60°, 75°, 90°, 105°, \ldots$$

The largest angle that is still acute (less than $90°$) is:
$$\mathbf{75°}$$ (which is $5 \times 15°$, formed between spokes that are 5 positions apart)

Note: $90°$ is a right angle (not acute), so the largest acute angle is $75°$.

Given conditions:
Let the angle be $x°$.

- $x$ is acute: 0° < x < 90° ... (1)
- $2x$ is acute: 0° < 2x < 90°, so x < 45° ... (2)
- $3x$ is acute: 0° < 3x < 90°, so x < 30° ... (3)
- $4x$ is acute: 0° < 4x < 90°, so x < 22.5° ... (4)
- $5x$ is obtuse: 90° < 5x < 180°, so 18° < x < 36° ... (5)

Combining all conditions:
From (4): x < 22.5°
From (5): x > 18°

So: 18° < x < 22.5°

Possibilities for $x$:
The angle $x$ must satisfy 18° < x < 22.5°.

If $x$ is a whole number of degrees: $x$ can be $\mathbf{19°, 20°, 21°, 22°}$.

Verification with $x = 20°$:
- $x = 20°$ ✓ (acute)
- $2x = 40°$ ✓ (acute)
- $3x = 60°$ ✓ (acute)
- $4x = 80°$ ✓ (acute)
- $5x = 100°$ ✓ (obtuse)

Answer: The possible whole-number degree measures are $\mathbf{19°, 20°, 21°, 22°}$ (any value strictly between $18°$ and $22.5°$).