Why enlarging the triangle leaves every ratio unchanged

What to assume they know

• Defining sine, cosine and tangent, then their three reciprocals — the six ratios and how each is built from two named sides
• Similar triangles, and the AA criterion, from Chapter 6 of this same book
• That in similar triangles corresponding sides are in the same proportion
• Multiplying a fraction top and bottom by the same non-zero number leaves it unchanged
• Recognising when two triangles share an angle because they share a pair of arms

What they should be able to do

• Explain why "the sine of A" needs justification before it can be used as notation
• Identify, in a figure where several right triangles stand on the same acute angle, which pairs are similar and by which criterion
• Use proportionality of corresponding sides to show that a chosen quotient takes the same value in a smaller and a larger triangle
• State the invariance carefully: what may change, what may not, and what the value does depend on
• Explain why a right triangle may be labelled k, 3k and so on rather than with actual lengths, and what k stands for
• Deduce the bound on sine and cosine from the hypotenuse being the longest side, and say why the invariance on its own cannot deliver it
• Run the argument backwards: from two acute angles having equal sines, conclude that the angles themselves are equal

Where it usually goes wrong

• "Obviously the ratios don't change — it's the same angle." That is the claim, not a reason for it. Until similarity is invoked, "same angle" and "same ratio" are two different statements, and the second is the one the chapter has to prove.
• "The triangles in Fig. 8.6 are similar because they look alike." They are similar because two angles agree. Appearance is what the picture supplies; the criterion is what the argument uses.
• "Bigger triangle, bigger sine." Bigger triangle, bigger sides — and the numerator and denominator grow by the same factor, so the quotient sits still. Show the 3-4-5 and 6-8-10 pair together.
• "k is the length of the side." k is whatever multiple you like. It is introduced precisely because the actual lengths are unknown and irrelevant, and it is guaranteed to vanish before any ratio is reported.
• "Since the ratio doesn't depend on the triangle, it doesn't depend on anything." It depends entirely on the angle. Move the angle and all six values move. The invariance is an invariance under scaling, not under everything.
• "Equal sines could come from different angles." Not among acute angles. Example 2 rules it out, which is why a sine value can be used to identify an angle at all.
• "AB = ±2√2·k means there are two triangles." The negative root is discarded because it would be a negative length, not because it is inconvenient.

Questions to check understanding

• Given a figure with nested right triangles on a common acute angle, state which are similar and why
• Show that a named ratio takes the same value in two similar right triangles
• Explain in one or two lines why a trigonometric ratio depends only on the angle
• Prove that two acute angles are equal, given that one of their ratios agrees — the Example 2 pattern, asked with sine, cosine or tangent
• Given a ratio, set up the sides using an unknown positive multiplier and complete the triangle with Pythagoras
• Justify the rejection of a negative square root when solving for a side

Examples worth working on the board

Inputs only. Values marked verified are worked out here on the chapter's printed data.

• Fig. 8.6 (§8.2, p. 116) — the figure the argument lives in. Angle A sits at the left. From A a horizontal arm runs right through M, then B, then N. From A a second arm slopes up through P, then C, then out to Q, the stretch beyond C drawn dotted. Three vertical segments drop from the sloping arm to the horizontal one: PM, then CB with the right-angle square drawn at B, then QN drawn dotted. The label Hypotenuse is printed alongside the arm AC. So the picture holds three right triangles nested on the same angle A: a small one PAM, the middle one CAB, and a large one QAN. These letters and the dotting are inside the artwork.
• The similarity step (§8.2, p. 116). Triangles PAM and CAB share the angle at A, and each has a right angle. Two angles match, so the AA criterion applies and the triangles are similar. The chapter names Chapter 6 as the source of the criterion — a deliberate backward cross-reference, outside this chapter's printed range.
• The proportion, and the cancellation (§8.2, pp. 116–117). Similarity gives one common value for AM/AB, AP/AC and MP/BC. Verified as an argument: call that common value t, so AM = t·AB, AP = t·AC and MP = t·BC. Then MP/AP = t·BC/(t·AC) = BC/AC, and the t is gone. The chapter states the resulting equalities for sine, then cosine, then tangent, and asks the reader to repeat the check in the large triangle QAN.
• The conclusion (§8.2, p. 117). Set in bold on the page: the values of these ratios do not shift when the triangle's sides are lengthened or shortened, so long as the angle is held. Note the conditional clause — it is doing real work, and section 12 exists to keep it in view.
• A numerical illustration. Take angle A with the legs 3 and 4 and hypotenuse 5, then double every length to 6, 8 and 10. Verified: sine is 4/5 in the first and 8/10 in the second, and 8/10 reduces to 4/5. Tangent is 4/3 and 8/6, which is again 4/3. Two triangles, different sizes, identical six numbers. Build this from the 3-4-5 shape that Example 1 on p. 118 produces, so the explanation is reusing a triangle the chapter has already drawn.
• The k habit in the chapter's own hands (§8.2, p. 117). Given a sine of 1/3, the chapter reads the statement as a proportion of two sides rather than as two lengths, sets BC = k and AC = 3k with k any positive number, and gets on with Pythagoras. Verified: AB² = (3k)² − k² = 9k² − k² = 8k², so AB = 2√2·k, and cos A = 2√2·k/(3k) = 2√2/3 — the k cancels, exactly as this topic predicts. The chapter attaches a bracketed Why? to the choice of the positive root; the answer is that k is a length and lengths are not negative.
• The bound (Remark, §8.2, p. 118). No leg of a right triangle can outrun the hypotenuse, so the two quotients that carry the hypotenuse underneath cannot exceed 1. Verified against the chapter's own later data: the largest value sine takes in Table 8.1 on p. 125 is exactly 1, at 90°. Note for accuracy: this Remark is worded on p. 118 as though the value were always strictly below 1, while the chapter summary on p. 132 says it never exceeds 1. The summary's form is the correct one.
• Example 2 (§8.2, pp. 118–119), the converse. Two right triangles, ABC with the right angle at C and PQR with the right angle at R, drawn side by side in Fig. 8.9, with the marked angles at B and at Q. The hypothesis is that the sines of those two angles agree. Verified as an argument: AC/AB = PR/PQ, so AC/PR and AB/PQ are equal — call the common value k. Pythagoras then recovers each remaining leg from the two sides already in hand — BC out of AB and AC in one triangle, QR out of PQ and PR in the other — and substituting AB = k·PQ and AC = k·PR pulls a factor k out of the square root, so BC/QR = k as well. All three pairs of corresponding sides are now in the same proportion, and the SSS-style similarity theorem the chapter cites as Theorem 6.4 delivers similar triangles, hence equal angles.
• The same argument to be run for cosine (Exercise 8.1 question 6, p. 121). The question asks for the equal-cosines version of Example 2. Verified: the proof is Example 2 with the adjacent leg in place of the opposite one throughout; nothing else changes.

Figures to have open

• Fig. 8.6 redrawn as a schematic: angle A, the two arms, the three perpendiculars at M, B and N, with the three triangles individually highlightable. This is the chapter's own figure and no substitute carries the argument as economically — the three triangles must be seen sharing the same angle before similarity is mentioned.
• A side-by-side of a 3-4-5 triangle and a 6-8-10 triangle at the same angle, with the six ratios computed under each and shown to agree. Standard schematic and the most persuasive single image in the topic.
• Fig. 8.9's pair of right triangles, right-angled at C and at R, with the marked angles at B and at Q. This is the chapter's own figure; the explanation needs the two right angles in the positions the book puts them, or the side names in the proof will not match.
• A cancellation panel: the scale factor written in both numerator and denominator and struck through. Standard schematic; it is the whole theorem in one frame.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 8 "Introduction to Trigonometry", §8.2, pp. 116–117 — Fig. 8.6, the similarity step, the proportionality equations and the boldface conclusion
• §8.2, p. 117 — the sine 1/3 reconstruction with Fig. 8.7, which the next topic takes over
• §8.2, p. 118 — the Remark bounding sine and cosine
• §8.2, pp. 118–119 — Example 2 with Fig. 8.9
• Exercise 8.1 question 6, p. 121 — the cosine form of Example 2
• Backward cross-references outside this chapter's printed pages: the AA criterion and the similarity theorem the book numbers 6.4 both belong to Chapter 6, and are cited here by chapter and theorem number only