Given one ratio, reconstructing the other five

What to assume they know

• Defining sine, cosine and tangent, then their three reciprocals — the six ratios and which two sides each one uses
• Why enlarging the triangle leaves every ratio unchanged — that the ratios track the angle and ignore the size, which is what makes an unknown multiplier safe
• Pythagoras theorem, and taking the square root of a squared length
• Simplifying a surd, and rationalising a denominator well enough to report an answer
• Solving a linear equation that arises from expanding a squared bracket

What they should be able to do

• Read a given ratio as a statement about the proportion of two named sides
• Assign lengths to those two sides using a single positive multiplier, and justify why any positive value will do
• Apply Pythagoras to obtain the third side in terms of the same multiplier
• Reject the negative square root with a reason rather than by habit
• Write all six ratios of the angle from the completed triangle, showing the multiplier cancelling
• Handle data given as a difference between two sides rather than as a ratio
• Evaluate a combination of ratios, such as a sum or difference of two squares, from the reconstructed triangle
• Judge whether a proposed value for a named ratio is possible at all

Where it usually goes wrong

• "Sine of A is 1/3 means the opposite side is 1 cm." It means the two sides stand in that proportion. The triangle has no determined size at all, which is precisely why the answer can be reported without one.
• "You have to find k." You never can and never need to. If k survives into an answer, the arithmetic has gone wrong somewhere.
• "The negative root is just ignored." It is rejected, and the reason is that it would be a negative length. The chapter marks this with a bracketed question rather than answering it.
• "Every problem starts from a ratio." Example 5 and Exercise 8.1 question 10 both start from a sum or difference of two sides, and are solved by expansion, not by a multiplier. Recognising which kind of data you have been given is half the skill.
• "A ratio bigger than 1 is impossible." Only for sine and cosine. Tangent, cotangent, secant and cosecant all exceed 1 routinely — Exercise 8.1 question 11 contains one of each kind on purpose.
• "The sum of the two squares in Example 3 came out as 1 by luck." It comes out as 1 for every angle, and §8.4 will prove it.
• "Three-four-five is a special trick." It is one shape among many. Inside the five pages this topic draws on, pp. 117–121, the reconstruct-from-a-ratio method produces 8-15-17 at Exercise 8.1 question 4 and 5-12-13 at question 5. Two more right triangles with whole-number sides turn up on the same pages by other routes: 20-21-29 in Example 3, where two of the lengths are simply given, and 5-12-13 again at question 10, where what is given is the sum of two sides. The method is the content, not the shapes.

Questions to check understanding

• Given one of the six ratios, find one, two or all of the others
• Given a ratio in the form "a whole number times a ratio equals a whole number", extract the ratio first and then proceed
• Given the sum or difference of two sides plus the third side, find all the ratios of a named angle
• Evaluate a combination such as a sum or difference of two squared ratios from a numerically specified triangle
• Check whether a stated equation between two expressions in the ratios holds, for a given value of one ratio
• Decide whether a proposed value of a named ratio is attainable, with justification

Examples worth working on the board

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

• The opening case (§8.2, p. 117, Fig. 8.7). Sine of A is given as 1/3. Fig. 8.7 is a right triangle with A at the lower left, the right angle at B, C above B, and the labels 3k on the hypotenuse and k on the vertical leg printed inside the artwork. Verified: AB² = 9k² − k² = 8k², so AB = 2√2·k, and cos A = 2√2/3. The remaining four follow: tan A = k/(2√2·k) = 1/(2√2), cosec A = 3, sec A = 3/(2√2), cot A = 2√2.
• Example 1 (§8.2, p. 118, Fig. 8.8). Tangent of A is given as 4/3. Fig. 8.8 has A at the lower left, the right angle at B at the lower right, C above B, with 3k printed along the base and 4k along the vertical leg. Verified: adding 16k² and 9k² gives 25k² for the square on the hypotenuse, so that side is 5k, and the six values read 4/5 for the sine, 3/5 for the cosine, 4/3 for the tangent as given, 5/4 for the cosecant, 5/3 for the secant and 3/4 for the cotangent. Worth naming: the 3-4-5 triangle, which will keep reappearing.
• Example 4 (§8.2, p. 120, Fig. 8.11). A right triangle ABC with the right angle at B, drawn with B at the lower left, C at the lower right and A above B, and the tangent at A given as 1. The task is to check that twice the product of the sine and cosine at A comes to 1. Verified: tangent 1 forces the two legs equal, so take both as k; then AC = k√2, and both the sine and the cosine equal 1/√2, whose product is 1/2, and twice that is 1. Note this quietly previews a double-angle fact, since the angle here is 45°.
• Example 3 (§8.2, p. 119, Fig. 8.10). Triangle ACB with the right angle at C, drawn with A at the top, C at the lower left, B at the lower right; the hypotenuse AB is marked 29 and the base CB is marked 21 inside the artwork, and the marked angle sits at B. Both are given as units, with no physical unit named. The two things asked for are the sum and the difference of the squares of the cosine and sine of that angle. Verified: AC² = 29² − 21² = (29 − 21)(29 + 21) = 8 × 50 = 400, so AC = 20. Then sine is 20/29 and cosine is 21/29. The sum of the squares is (400 + 441)/841 = 841/841 = 1. The difference is (441 − 400)/841 = 41/841. The first answer is a preview of §8.4.
• Example 5 (§8.2, p. 120, Fig. 8.12). Triangle OPQ with the right angle at P, drawn with Q at the top left, P at the lower left, O at the lower right, and 7 cm printed along PO inside the artwork. Given: OP = 7 cm, and OQ exceeds PQ by 1 cm. Verified: write OQ = 1 + PQ; then (1 + PQ)² = 49 + PQ², so 1 + 2·PQ = 49, giving PQ = 24 cm and OQ = 25 cm. Sine of Q is 7/25 and cosine of Q is 24/25. This is the case where the data is not a ratio.
• Exercise 8.1, the drill set (p. 121). Question 3 gives a sine of 3/4; verified: the third side is √7·k, so the cosine is √7/4 and the tangent is 3/√7. Question 4 gives fifteen times the cotangent equal to 8, that is a cotangent of 8/15; verified: the 8-15-17 triangle, so the sine is 15/17 and the secant is 17/8. Question 5 gives a secant of 13/12; verified: the 5-12-13 triangle, so cosine 12/13, sine 5/13, tangent 5/12, cotangent 12/5, cosecant 13/5. Question 7 gives a cotangent of 7/8 and asks for a quotient of two bracketed products and then the cotangent squared; verified: the first expression reduces to the cosine squared over the sine squared, which is the cotangent squared, so both answers are 49/64 — the point of the item is that they agree. Question 8 gives three times the cotangent equal to 4; verified: the 3-4-5 triangle again, and both sides of the proposed equation come to 7/25, so it does hold. Question 9 gives a tangent of 1/√3 in a triangle right-angled at B; verified: the two answers are 1 and 0, because the two acute angles add to 90° and the expressions are the sine and cosine of that sum. Question 10 puts the right angle at Q, sets PQ = 5 cm, and gives 25 cm as the total of PR and QR; verified: the same expansion trick as Example 5 gives QR = 12 and PR = 13, so sine of P is 12/13, cosine of P is 5/13, tangent of P is 12/5.
• The impossibility items (Exercise 8.1 question 11, p. 121). Five true-or-false claims: that a tangent is always below 1; that a secant can equal 12/5; that one abbreviation stands for another ratio's name; that an abbreviation is a product; and that a sine can equal 4/3. Verified: false, true, false, false, false. The two numerical ones turn on the same fact from opposite sides — a secant of 12/5 is above 1 and secants may be, while a sine of 4/3 is above 1 and sines may not be.

Figures to have open

• Fig. 8.7 and Fig. 8.8 redrawn as schematics, each labelled with the multiplier form the chapter prints inside the artwork — 3k and k for the first, 3k and 4k for the second. Anyone working only from the running text would not have these labels.
• Fig. 8.10 redrawn with 29 and 21 in the positions the book marks them and the angle marked at B. The right angle is at C, not at B, and the explanation's side names will come out wrong if that is moved.
• Fig. 8.12 redrawn with the right angle at P and 7 cm on the horizontal leg, since the problem's algebra depends on which side is the hypotenuse.
• A stack of three same-shape triangles at different sizes with identical ratios underneath. Standard schematic; it carries section 2 by itself.
• A four-box method loop: read the ratio, label with a multiplier, complete with Pythagoras, report the six. Standard schematic.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 8 "Introduction to Trigonometry", §8.2, p. 117 — the question this topic answers, and the sine 1/3 reconstruction with Fig. 8.7
• §8.2, p. 118 — the Remark bounding sine and cosine, and Example 1 with Fig. 8.8
• §8.2, p. 119 — Example 3 with Fig. 8.10
• §8.2, p. 120 — Example 4 with Fig. 8.11 and Example 5 with Fig. 8.12
• Exercise 8.1, p. 121, questions 1, 3, 4, 5, 7, 8, 9, 10 and 11, with Fig. 8.13
• Forward pointer inside this chapter: the sum of the two squares that Example 3 produces numerically is proved as an identity in §8.4, p. 128
• The chapter summary, §8.5, p. 132, point 3, states this topic's claim in one line