Recovering the angle between two lines from their two slopes

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What to assume they know

Steepness as the tangent of an angle, and the one line that has none — inclination, slope as its tangent, and the quotient form
Equal slopes mean parallel; slopes multiplying to minus one mean perpendicular — the two criteria of §9.2.2, and in particular that a product of slopes equal to minus one means a right angle
The tangent of a difference of two angles, from Chapter 3
That the tangent of an angle and the tangent of its supplement differ only in sign
Vertically opposite angles, and that two adjacent angles at a crossing fill a straight angle
Reading the sign of a quotient from the signs of its numerator and denominator

What they should be able to do

Explain why a crossing of two lines offers two candidate angles and why they are supplementary
Derive the tangent of the angle between two lines from the difference of their inclinations
State the condition the derivation attaches, and identify the geometric situation it rules out
Explain why that ruled-out situation costs nothing, by naming what §9.2.2 already says about it
Decide from the sign of the quotient which of the two angles the formula has produced
Write the acute-angle formula with its modulus and say what the modulus is doing
Recover the obtuse angle from the acute one
Given two slopes, compute both angles at the crossing
Given one slope and an angle, produce both admissible values of the second slope, and explain geometrically why there are two
Recognise when a problem's phrasing selects the acute angle and when it does not

Where it usually goes wrong

"Two lines make one angle." They make two, and the two are supplementary. Every difficulty in this section is a consequence of that, and the modulus at the end is the chapter's way of settling the ambiguity rather than dissolving it.
"The formula fails for perpendicular lines, so it is incomplete." It declines on precisely the case §9.2.2 settled with a product test. A formula that refused a case nobody could otherwise handle would be a defect; this one refuses the easiest case in the chapter.
"1 + m₁m₂ ≠ 0 is a technical condition about dividing by zero." It is a geometric condition wearing algebraic clothes. Read it aloud as "the lines are not perpendicular".
"An angle problem has one answer." Example 2 has two, and Fig 9.7 is printed to show why. Any problem that fixes an angle and one slope will generally admit two lines, one on each side.
"Take the bigger slope minus the smaller." The formula does not need an ordering. Swapping the two slopes negates the numerator, which swaps which of the two angles you have named — and under the modulus, changes nothing at all.
"A negative answer means a mistake." Before the modulus is applied, a negative value is information: it says the angle you happened to name is the obtuse one.
"The acute angle is always the one you want." It is the chapter's default, not a law. A question that specifies which angle, or asks for angles in the plural, overrides it.
"Vertical lines can be handled by taking a very large slope." They cannot be handled at all here. Both lines must have inclinations away from a right angle for their tangents to exist, which is why §9.2.3 opens by requiring both to be non-vertical.

Questions to check understanding

Given two slopes, find the acute angle between the lines and then the obtuse one
Given two lines in general form, extract slopes and find both angles
Given one slope and the angle, find both admissible values of the other slope and justify why two arise
Explain the condition attached to the formula in geometric language
Given three lines, decide which pair meets at the smallest angle
Find a line through a given point making a stated angle with a given line
Find the slope of a line equally inclined to two given lines

Examples worth working on the board

Inputs only. Values marked verified are worked out here on data printed inside pp. 156–174.

Fig 9.6 (§9.2.3, p. 156). Two lines drawn crossing above the x-axis: L₁ is the shallower of the two and crosses the x-axis to the left of the origin, L₂ is the steeper and crosses to the right. Read off the printed page: α₁ is arced at L₁'s crossing, α₂ at L₂'s, and at the intersection itself two angles are marked one above the other — θ in the wedge above and φ immediately below it. The figure is doing the work of the whole section, because it shows two named angles at one crossing rather than one.
The difference of inclinations (§9.2.3, p. 156). Verified: in the configuration drawn, the angle θ at the crossing is the amount by which the steeper line's inclination exceeds the shallower one's — the exterior-angle relation on the triangle cut off by the x-axis. Both inclinations must avoid a right angle for their tangents to exist at all, which is why both lines are required to be non-vertical.
The expansion (§9.2.3, p. 156). Verified: applying the tangent-of-a-difference identity to α₂ − α₁ and replacing tan α₁ by m₁ and tan α₂ by m₂ gives a quotient whose numerator is m₂ − m₁ and whose denominator is 1 + m₁m₂.
The condition (§9.2.3, p. 156, restated at p. 157 and again in the Summary, p. 174). Verified: the denominator 1 + m₁m₂ is zero exactly when m₁m₂ = −1, and by §9.2.2 that is exactly when the two lines are perpendicular. So the formula's one refusal is the one configuration whose angle needs no formula. Say this out loud — the chapter prints the condition four times over — twice on p. 156, once on p. 157 and once in the Summary — and never explains it once.
The supplementary partner (§9.2.3, p. 156). Verified: φ = 180° − θ, so tan φ is the negative of tan θ. The two angles at a crossing therefore always produce tangents of opposite sign, and only one of them can be positive.
The two cases (§9.2.3, p. 157). Verified: if the quotient is positive, the angle it names is acute and its partner is obtuse; if negative, the reverse. Since exactly one of the pair is acute whenever the lines are not perpendicular, the sign is a complete decision procedure and nothing else is needed.
The final formula (§9.2.3, p. 157). Read off the printed page — the printed expression carries modulus bars around the whole quotient, which the text layer does not preserve. Verified: taking the modulus discards the case distinction and returns the acute angle unconditionally; the obtuse one is then a subtraction from a straight angle.
Example 2 (p. 157). The angle between two lines is a quarter of a straight angle and one slope is 1/2; find the other. Verified: removing the modulus gives two equations. The branch equal to +1 yields m = 3; the branch equal to −1 yields m = −1/3. Both are genuine answers.
Fig 9.7 (p. 158). The picture the chapter supplies to explain why Example 2 has two answers. Read off the printed page: three lines are drawn and labelled with their slopes — the given line marked with slope 1/2 running up to the right, L₁ marked with slope −1/3, and L₂ marked with slope 3. Two arcs of 45° are marked, one at each of the two crossings. The value 8 is printed on the x-axis. Verified: the two answers are the two lines you reach by rotating the given line a quarter of a straight angle in the two available senses, and the product of the two answer slopes is −1, so the two answers are themselves perpendicular to each other. The chapter does not point that out.
Exercise 9.1 q10 — one line's slope is double another's, and the angle separating the two lines has tangent 1/3; find both slopes. Verified: writing the slopes as m and 2m and removing the modulus gives 2m² − 3m + 1 = 0 and 2m² + 3m + 1 = 0, so m ∈ {1, 1/2, −1, −1/2} and the admissible pairs are (1, 2), (1/2, 1), (−1, −2) and (−1/2, −1). Four pairs, not one — a good check on whether students remember the modulus.
Exercise 9.3 q8 — the angles between √3·x + y = 1 and x + √3·y = 1. Verified: the slopes are −√3 and −1/√3; the quotient's modulus is 1/√3, so the acute angle is 30° and its partner is 150°. The question asks for angles in the plural, which is the chapter being careful about exactly the point of this topic.
Exercise 9.3 q11 — two lines through (2, 3) meet at 60°, one of them has slope 2; find the other line. Verified: the two admissible slopes are (5√3 − 8)/11 and −(5√3 + 8)/11, approximately 0.060 and −1.515.
Miscellaneous Exercise q10 (p. 173) — lines through (3, 2) making 45° with x − 2y = 3. Verified: the given line has slope 1/2, so this is Example 2 again with a point attached, and the two slopes are 3 and −1/3. Worth pairing with Example 2 explicitly.
Miscellaneous Exercise q18 (p. 173) — the two lines y = 3x + 1 and 2y = x + 3 each make the same angle with y = mx + 4; find m. Verified: equating the two moduli and taking the branch that has real solutions gives 7m² − 2m − 7 = 0, so m = (1 ± 5√2)/7, approximately 1.153 and −0.867. The other branch produces m² = −1 and no real line, which is itself instructive.
Miscellaneous Exercise q23 (p. 174) — a person standing where 2x − 3y + 4 = 0 meets 3x + 4y − 5 = 0 must get to the path 6x − 7y + 8 = 0 as quickly as possible. Data handed over intact; the reasoning that the quickest route is the perpendicular one belongs to The shortest route from a point to a line, read off its coefficients, but the crossing point is located here.

Figures to have open

Fig 9.6 redrawn, with both inclinations arced at the x-axis and both adjacent angles marked at the crossing. The chapter's own figure, and the section cannot open without two angles visible at one crossing.
Fig 9.7 redrawn, all three lines labelled with their slopes and both 45° arcs kept. The chapter's own figure; it is the chapter's answer to "why two answers".
A movement of one line rotating about a fixed point with the quotient displayed, showing the value change sign as the line passes through the perpendicular direction. Standard schematic; it makes the excluded case visible as a crossing rather than a rule.
A pair of crossings side by side, one with a positive quotient and one with a negative one, each annotated with which of the two angles the value named. Standard schematic built from the chapter's two cases.

Where this sits in the book

NCERT Mathematics, Textbook for Class XI, Chapter 9 "Straight Lines", §9.2.3 Angle between two lines, pp. 156–157 — Fig 9.6, the derivation, the condition, both sign cases and the acute-angle formula
Example 2, p. 157, with Fig 9.7 on p. 158 explaining its two answers
Exercise 9.1, p. 159, question 10
Exercise 9.3, p. 167, question 8, and p. 168, question 11
Miscellaneous Exercise on Chapter 9, pp. 173–174, questions 10, 18 and 23
The chapter Summary, p. 174 — the acute-angle formula with its condition restated
Cross-reference outside this chapter: the tangent of a difference of two angles is a Chapter 3 identity, used here without re-derivation