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.