Squeezing 30°, 45° and 60° out of two special triangles

What to assume they know

• Defining sine, cosine and tangent, then their three reciprocals — the six ratios and which sides each uses
• Given one ratio, reconstructing the other five — reading a shape with an unknown multiplier and completing it with Pythagoras
• The angle sum of a triangle, and that equal angles face equal sides
• That every angle of an equilateral triangle is 60°
• Congruent triangles and the CPCT step — corresponding parts of congruent triangles are equal
• Simplifying surds well enough to write 1/√2 and √3/2 confidently

What they should be able to do

• Explain why exact values are available for some angles and not for others
• Construct the isosceles right triangle and derive all six ratios of 45° from it
• Construct the half of an equilateral triangle and derive all six ratios of 30° and of 60° from it
• Justify why the perpendicular from the apex of an equilateral triangle bisects both the base and the apex angle
• Use the derived values to find an unknown side of a right triangle from one side and one angle
• Use the derived values to find an unknown angle from two sides
• Solve a pair of simultaneous conditions on a sum and a difference of two angles

Where it usually goes wrong

• "These three values were measured and tabulated." They were derived.
• "The 30-60-90 triangle needs its own construction." It does not — it is half an equilateral triangle, and that is where the exact halving of the hypotenuse comes from. If a student cannot say where the factor of two came from, they have memorised rather than understood.
• "The perpendicular obviously bisects the base." It follows from congruence, and the chapter marks the step with a bracketed question. Answer it: two equal sides, a shared side, and a right angle each.
• "1/√2 and √2/2 are different answers." They are the same number written two ways. Pick one form and stay with it; boards accept both.
• "The 30° and 60° columns are two separate things to learn." They are one triangle read from its two ends. Learning one and the swap gives the other for free.
• "tan 45° = 1 because 45 is halfway." It equals 1 because the two legs are equal. Nothing about the number 45 is doing the work.
• "You can just read the answer off the table." For Example 6 you must first decide which ratio connects the side you have to the side you want. Choosing the ratio is the skill; the table is the lookup afterwards.

Questions to check understanding

• Evaluate a numerical expression built from the ratios of the special angles
• Multiple-choice items where an expression must be recognised as a single ratio of a special angle
• Find an unknown side of a right triangle given one side and one special angle
• Find the angles of a right triangle given two of its sides
• Given the sine or cosine of a sum and of a difference of two angles, solve for both
• Derive the value of a named ratio at 30°, 45° or 60° from a construction, with the triangle drawn as part of the answer

Examples worth working on the board

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

• The framing (§8.3, p. 121). The section opens by recalling that 30°, 45°, 60° and 90° are angles the student can already construct, and announces that it will also handle 0°. The four constructible angles are the reason the list is the length it is.
• Fig. 8.14 and the 45° case (§8.3, p. 122). A right triangle lettered ABC, its right angle at B, drawn with A at the lower left, B at the lower right and C directly above B; no lengths appear on the artwork. Given that one acute angle is 45°, the other must be too, since the two acute angles of a right triangle add to 90°. Equal angles face equal sides, so the two legs are equal; call each of them a. Verified: AC² = a² + a² = 2a², so AC = a√2. Then the sine and cosine of 45° are both a/(a√2) = 1/√2, the tangent is a/a = 1, the cosecant and secant are both √2, and the cotangent is 1.
• Fig. 8.15 and the 30°/60° case (§8.3, pp. 122–123). An equilateral triangle ABC drawn with A at the apex, B at the lower left, C at the lower right, and the perpendicular AD dropped to the base, with D between B and C. Two angle values are printed inside the artwork: 30° at the apex, marking the half-angle at A, and 60° at B. Verified as an argument: the two halves ABD and ACD have AB = AC because the triangle is equilateral, share the side AD, and each has a right angle at D, so they are congruent; CPCT then gives BD = DC and splits the apex angle into two equal 30° pieces. So ABD is a right triangle with a 30° angle at A and a 60° angle at B.
• The half-triangle measured (§8.3, p. 123). Set AB = 2a. Then BD is half of BC, and BC equals AB because the triangle is equilateral, so BD = a. Verified: AD² = (2a)² − a² = 4a² − a² = 3a², so AD = a√3.
• The twelve values, all read off that one picture (§8.3, p. 123). Verified by reading BD, AD and AB against the two angles in turn. At 30°: sine 1/2, cosine √3/2, tangent 1/√3, cosecant 2, secant 2/√3, cotangent √3. At 60°: sine √3/2, cosine 1/2, tangent √3, cosecant 2/√3, secant 2, cotangent 1/√3. Note: the 30° list and the 60° list are each other's read in reverse, because the two angles are the two acute angles of one triangle and their legs swap — the same swap the opening topic set up.
• Example 6 (§8.3, pp. 125–126, Fig. 8.19). A triangle lettered ABC whose right angle sits at B, drawn with A at the top left, B beneath it and C at the lower right; AB = 5 cm is printed alongside the vertical leg and 30° is printed at C, inside the artwork. Verified: the tangent of 30° is AB/BC, so 5/BC = 1/√3 and BC = 5√3 cm; the sine of 30° is AB/AC, so 5/AC = 1/2 and AC = 10 cm. The page then checks the second answer with Pythagoras — 25 + 75 = 100 — which is worth showing, because it demonstrates that the two routes agree.
• Example 7 (§8.3, p. 126, Fig. 8.20). Triangle PQR right-angled at Q, drawn with P at the top left, Q at the lower left and R at the lower right; 3 cm is printed alongside PQ and 6 cm alongside PR, inside the artwork. Verified: the sine of the angle at R is 3/6 = 1/2, so that angle is 30°, and the angle at P is therefore 60°. The page closes with the useful generalisation that one side plus any one other part is enough to pin down the whole right triangle.
• Example 8 (§8.3, p. 126). Given that the sine of the difference of two angles is 1/2 and the cosine of their sum is 1/2, with the sum lying above 0° and not above 90°, and the first angle the larger. Verified: the difference is 30° and the sum is 60°, so the two angles are 45° and 15°. The range condition is what makes the two readings unique.
• Exercise 8.2, the evaluation set (p. 127). Hand the data over intact. Question 1 has five expressions to evaluate; verified, in order: (i) 1; (ii) 2; (iii) (3√2 − √6)/8; (iv) (43 − 24√3)/11; (v) 67/12. Question 2 is four multiple-choice items; verified: (i) is the sine of 60°, (ii) is 0, (iii) holds at 0°, (iv) is the tangent of 60°. Note that items (i) and (iv) differ only in the sign in the denominator and land on different answers — that is the point of printing both. Question 3 gives a tangent of √3 for the sum and 1/√3 for the difference, under the same range condition as Example 8; verified: the sum is 60°, the difference 30°, so the angles are 45° and 15°.

Figures to have open

• Fig. 8.14 redrawn with the two legs marked a and the hypotenuse a√2, and both acute angles marked 45°. Standard schematic; the chapter's own version carries no lengths.
• Fig. 8.15 redrawn as a movement: the full equilateral triangle, then the perpendicular appearing, then the two halves separating and one of them being labelled 2a, a and a√3. This is the chapter's own figure and the derivation cannot be shown without it. The printed version marks 30° at the apex and 60° at the base vertex, inside the artwork.
• The finished value table for 30°, 45° and 60°, built column by column as each is derived rather than shown complete. The chapter's own Table 8.1 on p. 125 also carries 0° and 90°, which belong to the next topic.
• Fig. 8.19 and Fig. 8.20 redrawn with the lengths and angles the book prints inside their artwork — 5 cm and 30° for the first, 3 cm and 6 cm for the second.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 8 "Introduction to Trigonometry", §8.3 Trigonometric Ratios of Some Specific Angles, p. 121 — the framing and the list of angles
• §8.3, p. 122 — the 45° subsection with Fig. 8.14, and the opening of the 30°/60° subsection with Fig. 8.15
• §8.3, p. 123 — the half-triangle measured, and the twelve values for 30° and 60°
• §8.3, p. 125 — Table 8.1, whose middle three columns this topic derives, and Example 6 with Fig. 8.19
• §8.3, p. 126 — Example 7 with Fig. 8.20, and Example 8
• Exercise 8.2, p. 127, questions 1, 2 and 3
• The chapter summary, §8.5, p. 132, point 4, names the five angles whose values are expected