Chapter 11.2 — Exercise 11.1 — Trigonometric Ratios

Four Direct Applications

The last two rows share a technique worth naming: when only a ratio is given rather than actual lengths, introduce a scaling constant k for the two sides that ratio compares, then let Pythagoras find the third side in terms of that same k. The k always cancels away by the time a final ratio is computed, which is exactly why it never needs to be solved for explicitly.

The second row is worth a second look too, since it asks for a difference of two ratios rather than a single one. tanP and tanR are reciprocals of each other here — both built from the same pair of legs, just inverted — so the subtraction genuinely doesn't simplify away to anything cleaner than the fraction 527/168 it lands on. Not every trigonometric expression is designed to collapse to a tidy whole number; some are simply asking for the arithmetic to be carried through correctly.

A Correction Worth Flagging

△ABC has a right angle at B, with BC=24 and AC=25. Finding AB first: AB²=625−576=49, so AB=7.

It's worth substituting carefully here specifically because tanθ and cosθ share the same denominator-looking number (25) only in cosθ's case — tangent never involves the hypotenuse at all, only the two legs. Confusing which side plays which role in each of the three basic ratios is the single easiest mistake to make once a triangle has three different side lengths in play at once.

A reliable habit worth building here: write out the three side lengths first — hypotenuse, opposite, adjacent, relative to whichever angle the problem asks about — before substituting anything into a ratio formula. Reaching straight for a formula and filling in numbers from memory, without first pinning down which length is which, is exactly how a value from one ratio's calculation quietly leaks into a different ratio's answer.

A Proof With No Numbers at All

Given two entirely separate right triangles ABC and XYZ where cosA=cosX, prove ∠A=∠X. Writing AB/AC = XY/XZ = k for both triangles, Pythagoras gives BC/YZ = k as well — meaning all three side ratios between the two triangles match exactly, one after another, making them similar. Similar triangles share every angle, so ∠A=∠X follows immediately.

This proof never touches a single actual measurement — it works for every possible pair of right triangles satisfying the one given condition, precisely because it argues from the ratio definitions themselves rather than from any specific numbers substituted into them.

This result is worth stating as a general fact in its own right, not just as the answer to one specific problem: within the range of angles a right triangle can actually contain (strictly between 0° and 90°), cosine is one-to-one — no two different angles in that range ever share the same cosine value. That's exactly what makes "cosA=cosX implies A=X" true in the first place, and the identical reasoning applies equally well to sine and to tangent over the same range.

A Ratio That Doesn't Simplify to a Whole Number

One final variation: given cotθ=7/8 (so the adjacent side is 7k and the opposite side is 8k), the hypotenuse comes out irrational this time: AC=√(49k²+64k²)=√113 k — a genuinely different situation from every problem before it, since 113 isn't a perfect square and no amount of careful arithmetic makes the square root disappear.

Both parts substitute the identical two ratio values found above into two differently-shaped expressions, and it's worth predicting before calculating which one is more likely to simplify cleanly — an expression built entirely from squares tends to invite the sin²+cos²=1 identity, while one left as a plain unsquared sum usually doesn't.

Part (i) is worth noticing for how cleanly it resolves despite the irrational sine and cosine values — every √113 cancels out completely once the expression is rewritten in terms of cot²θ, leaving a plain fraction of whole numbers. Not every trigonometric expression built from irrational ratios simplifies this cleanly; recognising the (1−sin²θ)/(1−cos²θ) pattern as secretly being cos²θ/sin²θ is what makes it possible here.

Part (ii), by contrast, genuinely can't shed its surd — the final answer (√113+8)/7 stays irrational no matter how it's rearranged, since nothing in the expression pairs up the way it did in part (i) to cancel the root away. Both parts start from the identical two ratios, sinθ=8/√113 and cosθ=7/√113, which is worth noticing: whether a surd survives to the final answer or cancels out entirely depends entirely on the specific shape of the expression being evaluated, not on whether the starting ratios themselves happen to be irrational.

A Preview of the Next Two Exercises

Given tanA=√3 in a right triangle (BC=√3 k, AB=k), the hypotenuse is AC=2k — recognisably the 30°-60°-90° triangle, giving A=60° and C=30°. Spotting this specific triangle shape by its ratio alone, before doing any further calculation, is a skill worth building now, since it comes up constantly once specific angle values are introduced properly in the next exercise.

Both results are quietly previewing ideas the next two exercises make explicit: the specific values √3/2, 1/2, and so on come directly from the standard 30°/60° angle table in Exercise 11.2, and A+C=90° here is exactly the complementary relationship Exercise 11.3 builds into a general rule.

It's worth noticing that both computed results, 1 and 0, are themselves recognisable trigonometric values — sin90° and cos90° respectively, since A+C=90° here. That's not a coincidence specific to this one triangle: sinA·cosC+cosA·sinC is always equal to sin(A+C) for any two angles, a genuine identity that this problem's specific numbers happen to verify rather than derive from scratch.

From One Angle to Five Standard Ones

Every ratio in this exercise came from an angle whose exact value was never actually named — only its sine, cosine, or tangent, computed from raw side lengths. Exercise 11.2 flips that around, starting from five specific, nameable angles and deriving their exact ratio values once and for all, a genuinely different direction of travel from every problem worked through in this exercise. For the six ratio definitions this exercise applied throughout, and the Pythagorean-triple shortcuts worth watching for in every problem, revisit the chapter introduction.