Picking the one ratio that links what you know to what you want

What to assume they know

• The sightline, and the horizontal it gets measured against — turning a scene into a lettered right triangle
• The definitions of sine, cosine, tangent and their reciprocals in terms of the opposite side, the adjacent side and the hypotenuse
• Exact values at 30°, 45° and 60°, and that tan 45° is 1
• Similar triangles and the equality of corresponding side ratios, from Chapter 6
• Rearranging a proportion to isolate either the numerator or the denominator
• Rationalising a denominator, and substituting a decimal approximation for √3

What they should be able to do

• Letter a right triangle so that the known side, the wanted side and the known angle are all identifiable at a glance
• Classify each of a triangle's three sides as opposite, adjacent or hypotenuse relative to a stated acute angle
• Select the ratio determined by the (known side, wanted side) pair, and state why no other ratio would serve
• Explain, from similarity, why the ratio depends only on the angle and not on the triangle's size
• Show that a ratio and its reciprocal give the same equation, and choose between them on convenience rather than correctness
• Solve for the wanted side when it appears in the numerator, and when it appears in the denominator
• Take a second measurement from the same figure using a second ratio, without redrawing
• Keep an exact surd answer and, separately, report a decimal one using a stated approximation, saying which is which

Where it usually goes wrong

• "Always use tan." Example 2 wants the ladder itself, so the pair is opposite-and-hypotenuse and the answer is sin. Reaching for the tangent there produces a length that is not the ladder.
• "Opposite and adjacent are properties of the sides." They are properties of a side relative to a chosen angle. In Fig. 9.5 the segment BD is opposite the 60° at C; relative to the angle at B it would be adjacent.
• "The hypotenuse is the bottom side / the long side of the drawing." It is the side facing the right angle, wherever the drawing puts it. In Fig. 9.5 the hypotenuse is the ladder and it is drawn leaning.
• "tan and cot are two different methods." They are one equation written the two ways round. Choose whichever puts the unknown where it is easiest to isolate.
• "The ratio changes if the triangle is bigger." It does not, and that is the entire reason the method works. Show two nested right triangles at 60° and compute the quotient in both.
• "√3 = 1.73, so the answer is exact." The chapter itself writes its abbreviation for "approximately" on the results of Example 2 and leaves Example 1's answer as a surd. Two different reporting conventions one page apart — Example 1 finishes on p. 135 and Example 2 runs on p. 136, which in a bound copy is the far side of the same leaf, not the opposite half of a spread. Whether the book meant anything by the switch it does not say.
• "Once you have used a ratio the figure is finished with." Example 2 asks a second question of the same figure and needs a second ratio.

Questions to check understanding

• Name the ratio that connects two stated sides of a lettered right triangle, with no numbers supplied
• Compute a height, a horizontal distance or a slant length from one angle and one length
• The same problem asked twice of one figure, wanting two different sides
• A problem whose answer is the sum of two computed pieces, such as a broken tree or a chimney above an observer
• Report an answer both as a surd and to two decimal places from a given approximation for √3
• Explain why a trigonometric ratio can be tabulated in advance of any particular triangle

Examples worth working on the board

Values marked verified are worked out here on data printed inside pp. 133–143.

• The chapter's own statement of the rule (§9.1, p. 135). Working on Fig. 9.1, the page asks in so many words which ratio holds the two lengths already in hand together with the one still wanted, narrows to the tangent and the cotangent because those are the two that involve AB and BC, and writes both down: tan A = BC/AB and cot A = AB/BC. That narrowing sentence is the single most transferable line in the chapter.
• The three pairings, as a table for section 1 (mine, from the definitions): opposite with adjacent → tangent or cot; opposite with hypotenuse → sin or its reciprocal; adjacent with hypotenuse → cosine or its reciprocal. Six names, three pairs, because every ratio has a reciprocal that carries the same information.
• Example 1 (p. 135, Fig. 9.4). Inputs: a vertical tower; a ground point 15 m from its foot; elevation 60° there. Fig. 9.4 letters the top A, the foot B, the ground point C, prints 60° at C and marks 15 m along CB. Known side CB is adjacent to the 60°; wanted side AB is opposite it. Verified: AB = 15 tan 60° = 15√3 m ≈ 25.98 m. The page leaves the surd standing.
• Example 2 (p. 136, Fig. 9.5). Inputs: a pole 5 m tall; the electrician must reach a point 1.3 m below its top; the ladder is to stand at 60° to the ground. She is told to take √3 = 1.73. Wanted: the ladder's length, and how far from the pole its foot goes. Fig. 9.5 letters the pole's top A, the reach point B, the pole's foot D, the ladder's foot C, and prints 60° at C; the ladder BC is drawn as a rung ladder against the pole.
  • The reach height comes off by subtraction first: BD = 5 − 1.3 = 3.7 m. This is not trigonometry.
  • Wanted side BC is the hypotenuse and known side BD is opposite the 60°, so the pair names sin. Verified: BC = 3.7/sin 60° = 7.4/√3 ≈ 4.28 m using the chapter's own √3 = 1.73.
  • Second question, same triangle: wanted DC is adjacent, known BD is opposite, so the pair names cot. Verified: DC = 3.7/√3 ≈ 2.14 m.
  • The teaching point of this example is that the figure did not change between the two questions — only the wanted side did, and the ratio changed with it. Build section 7 on exactly this.
• Example 3 (p. 137, Fig. 9.6). Inputs: an observer 1.5 m tall, 28.5 m from a chimney, reading 45° to its top. Verified: the tangent gives AE = 28.5 tan 45° = 28.5 m, and the chimney is AB = 28.5 + 1.5 = 30 m. The ratio answered only the triangle; the last metre and a half came from the figure. Worth pairing with Example 1: there the ratio was the whole answer, because the sighting was from ground level.
• A reciprocal-pair demonstration for section 5 (mine). Take Example 1 and solve it a second time as cot 60° = 15/AB, i.e. AB = 15/cot 60° = 15√3. Same number, one extra reciprocal. The point is that the offer of two ratios on p. 135 is not a choice between two methods.
• Exercise 9.1 items that are pure single-ratio selection (pp. 141–142), inputs only. Q1 (Fig. 9.11): a taut rope 20 m long from a vertical pole's top to the ground, meeting the ground at 30°; the figure letters the pole's top A, its foot B and the ground anchor C, prints 30° at C and marks 20 m along the rope. Q2: a storm breaks a tree so the top bends over and touches the ground 8 m from the trunk's foot, making 30° there; the tree's original height is wanted. Q3: two slides, one with its top 1.5 m up at 30° to the ground, one with its top 3 m up at 60°; each slide's length is wanted. Q4: elevation 30° from a ground point 30 m from a tower's foot. Q5: a kite 60 m up on a slack-free string at 60° to the ground. Verified, working added here: Q1 → 20 sin 30° = 10 m. Q2 → the broken piece is 8/cos 30° = 16/√3 and the standing stump is 8 tan 30° = 8/√3, so the tree was 24/√3 = 8√3 ≈ 13.86 m; note the answer is a sum of two pieces, which is the trap. Q3 → 1.5/sin 30° = 3 m and 3/sin 60° = 2√3 ≈ 3.46 m. Q4 → 10√3 ≈ 17.32 m. Q5 → 40√3 ≈ 69.28 m.

Figures to have open

• Fig. 9.4 (p. 135): tower AB, ground point C, 15 m marked, 60° at C. Standard schematic.
• Fig. 9.5 (p. 136): the pole with A on top, B at the reach point, D at the foot, the ladder from C leaning against B, 60° at C. The 5 m and the 1.3 m are in the text, not on the drawing.
• Fig. 9.6 (p. 137): chimney AB, observer CD, eye-level point E, 45° at D.
• Fig. 9.11 (p. 141): pole AB, anchor C, 20 m marked along the rope, 30° at C. Standard schematic.
• A purpose-built similarity figure for section 3: two right triangles with a common 60° angle, one drawn inside the other, with the corresponding sides colour-matched. This is an added figure; the chapter assumes similarity rather than drawing it.

Where this sits in the book

• NCERT Class X Mathematics, Chapter 9 "Some Applications of Trigonometry", §9.1 "Heights and Distances": the ratio-selection argument on p. 135, then Examples 1, 2 and 3 on pp. 135–137.
• Figs. 9.4 (p. 135), 9.5 (p. 136), 9.6 (p. 137), 9.11 (p. 141).
• Exercise 9.1 questions 1 to 5, pp. 141–142.
• §9.2 Summary, p. 143, point 2 — the one line in which the chapter says what the ratios are for.
• Chapter 6 is where similarity was established; this chapter uses it silently and the citation is a deliberate cross-reference outside pp. 133–143.