Carry trig back to any triangle — convert freely between sides and angles.
For seven lessons trigonometry has lived first inside a right triangle, then on the unit circle, then as a wave, then as an algebra of identities. Now it comes home to where it began — an ordinary triangle — but this time with no right angle to lean on. Given a few sides and angles, can we always recover the rest? The answer is two short equations. The law of sines pairs each side with the sine of its opposite angle; the law of cosines is the Pythagorean theorem with a correction term for the tilt. Between them they solve any triangle and even tell us how many triangles a given set of clues allows — so we can measure a river we cannot cross or a mountain we cannot climb.
Label a triangle the usual way: side a sits opposite angle A, side b opposite B, side c opposite C. The longest side always faces the largest angle — push an angle wider and the side across from it stretches. The law of sines makes that intuition exact: the ratio of a side to the sine of its opposite angle is the same for all three corners.
asin A = bsin B = csin C = 2R
Why is the common value 2R, twice the radius of the circle through all three corners? Drop the triangle into its circumscribed circle: a chord of length a subtends an inscribed angle A, and the inscribed-angle relationship from Stage 18 forces a = 2R sin A. The same holds for every side, so all three ratios collapse to the one number 2R.
This is the tool for AAS and ASA — clue sets where you know an angle and the side across from it. Find the third angle from A + B + C = 180°, then read any missing side straight off a known side–angle pair.
A triangle has A = 40°, B = 75°, and the included side c = 6. Then C = 180° − 40° − 75° = 65°. The law of sines gives a = c · sin A / sin C = 6 · sin 40° / sin 65° ≈ 6 · 0.643 / 0.906 ≈ 4.26, and b = 6 · sin 75° / sin 65° ≈ 6.40. The triangle is solved.
The law of sines links a side to the angle opposite it. Use it the moment you can pair a known side with its own opposite angle — that is exactly the AAS and ASA situations, and (with care) the SSA situation in §25.7.3.
The law of sines is helpless when you know two sides and the angle between them (SAS), or all three sides (SSS) — in neither case can you pair a side with its opposite angle yet. For those we need the law of cosines, which generalizes Pythagoras to any angle at the corner:
c2 = a2 + b2 − 2ab cos C
Read it as Pythagoras plus a tilt-correction. When the included angle C = 90°, cos 90° = 0 and the term −2ab cos C vanishes, leaving c2 = a2 + b2 — exactly the Pythagorean theorem. Open the corner past 90° and cos C turns negative, so the correction adds length and c grows. Close it below 90° and cos C is positive, so the correction subtracts and c shrinks. The cosine measures exactly how far the corner has tilted from square.
Sides a = 6, b = 8, c = 9. Solve the cosine law for an angle: cos C = (a2 + b2 − c2)/(2ab) = (36 + 64 − 81)/96 = 19/96 ≈ 0.198, so C ≈ 78.6°. Repeat for A and B, or finish with the law of sines: the three angles come out to about A ≈ 40.8°, B ≈ 60.6°, C ≈ 78.6° — and they sum to 180°, as they must.
Two laws, four standard clue sets. Have a side with its opposite angle? Use the law of sines (AAS, ASA). Have the angle between two sides, or all three sides? Use the law of cosines (SAS, SSS). One of the two will always open the door.
There is one clue set that refuses to behave: SSA — two sides and a non-included angle (an angle not between them). Suppose you know angle A, the side b next to it, and the side a across from it. You can try the law of sines — but the picture warns you that the answer may not be unique.
Here is the geometry. Lay down angle A with the long ray of length b. From the far end of that ray, swing side a like a compass and see where it lands on the base. The shortest reach from that pivot to the base is the perpendicular, of height h = b sin A. Everything turns on how a compares to that height:
| condition (A acute) | swing arc hits the base… | triangles |
|---|---|---|
| a < h = b sin A | not at all | 0 |
| a = h | once, at the foot (right angle) | 1 |
| h < a < b | in two places | 2 |
| a ≥ b | once (the other crossing is behind A) | 1 |
The two-triangle case is the famous trap. When h < a < b the arc cuts the base on both sides of the foot of the perpendicular, giving an acute angle B and its supplement 180° − B — two genuinely different triangles from the same three numbers.
Never assume SSA has one answer. The law of sines hands you B = sin−1(b sin A / a), but its supplement 180° − B may also work. Always test against h = b sin A first: if a < h there is no triangle; if h < a < b there are two. Drawing the swing arc settles it every time.
Area in middle school was ½ × base × height. But the height is awkward when there is no right angle. Trig supplies it for free: with sides a and b meeting at angle C, the height onto base b is a sin C, so
Area = ½ · b · (a sin C) = ½ab sin C
Two sides and the angle between them — the SAS clue set again — give the area outright. When instead you know all three sides, Heron's formula finishes the job: Area = √(s(s−a)(s−b)(s−c)) with the semiperimeter s = (a+b+c)/2. The two formulas must agree, and that agreement is a fine arithmetic check.
For a = 5, b = 7, C = 40°: Area = ½ · 5 · 7 · sin 40° ≈ 17.5 · 0.643 ≈ 11.25. The third side is c ≈ 4.50, so the semiperimeter is s ≈ 8.25 and Heron gives √(8.25·3.25·1.25·3.75) ≈ √126.5 ≈ 11.25. The two agree, as they must.
The whole reason trigonometry left the right-triangle cage is to measure things we cannot reach. The recipe never changes: sketch the triangle, label what you know, pick the law that fits, solve.
Pace off a baseline PQ = 120 m along the bank. Sight a tree T on the far side: the angle at P is 63°, at Q it is 51°. Then the angle at T is 180° − 63° − 51° = 66°. By the law of sines, PT = 120 · sin 51° / sin 66° ≈ 120 · 0.777 / 0.914 ≈ 102 m. The river's width is PT · sin 63° ≈ 91 m — measured without a single step into the water.
A mountain's height works the same way: two elevation angles taken a known distance apart give an ASA (or AAS) triangle for the slant distance, then a sine for the vertical. A ship's bearing problem is a triangle whose corners are two landmarks and the ship; the law of cosines turns two known legs and the turn angle between courses into the straight-line distance home. Whenever the figure is a triangle and you can name a side-with-its-angle or an angle-between-two-sides, one of the two laws solves it.
Match the clue set to the law. Whenever you can pair a side with its opposite angle, reach for the law of sines; whenever you have the angle between two sides, or all three sides, reach for the law of cosines. SSA is the one to watch — always count first.
| you know… | clue set | law to use | note |
|---|---|---|---|
| all three sides | SSS | cosine | solve cos for each angle |
| two sides + the angle between | SAS | cosine | third side, then angles |
| two angles + any side | ASA / AAS | sine | always one triangle |
| two sides + a non-included angle | SSA | sine + count | 0, 1, or 2 — test a vs h = b sin A |
asin A = bsin B = csin C = 2R · c2 = a2 + b2 − 2ab cos C · Area = ½ab sin C
That toolbox of measurable relationships — sides and angles you can convert into one another at will — is exactly what the next strand needs. Stage 26 · Sequences begins by watching quantities change in regular steps, and many of those patterns first show up as the side or angle of a triangle that grows by a fixed rule.
In triangle ABC, A = 50°, B = 60°, and side a = 8. Find side b.
Side and opposite angle are paired → law of sines. b = a · sin B / sin A = 8 · sin 60° / sin 50° ≈ 8 · 0.866 / 0.766 ≈ 9.04.
A triangle has sides a = 7 and b = 10 with included angle C = 60°. Find the third side c.
Two sides + the angle between → law of cosines. c2 = 72 + 102 − 2·7·10·cos 60° = 49 + 100 − 140·0.5 = 79, so c = √79 ≈ 8.89.
Find the largest angle of the triangle with sides 4, 5, 6.
The largest angle faces the longest side (6). With a = 4, b = 5, c = 6: cos C = (16 + 25 − 36)/(2·4·5) = 5/40 = 0.125, so C = cos−1(0.125) ≈ 82.8°.
For A = 30°, b = 8, decide how many triangles exist when (i) a = 3, (ii) a = 4, (iii) a = 5.
The height is h = b sin A = 8 · sin 30° = 4. (i) a = 3 < h = 4 → 0 triangles. (ii) a = 4 = h → 1 (a right angle at the foot). (iii) h = 4 < a = 5 < b = 8 → 2 triangles.
Find the area of the triangle with a = 9, b = 12, and included angle C = 50°, then check it against Heron's formula.
Area = ½·9·12·sin 50° = 54·0.766 ≈ 41.4. Check: the third side is c = √(81 + 144 − 2·9·12·cos 50°) ≈ √86.2 ≈ 9.28, so s ≈ 15.14 and Heron gives √(15.14·6.14·3.14·5.86) ≈ √1712 ≈ 41.4 — agreement.
Two roads leave a town at an angle of 70°. One car drives 40 km along one road, another drives 55 km along the other. How far apart are they?
Two sides + the angle between → law of cosines. d2 = 402 + 552 − 2·40·55·cos 70° = 1600 + 3025 − 4400·0.342 ≈ 3120, so d ≈ √3120 ≈ 55.9 km.
Six questions to lock it in. Tap the answer you think is right.
This lesson covers the Common Core geometry standards G-SRT.D.10 (prove the laws of sines and cosines) and G-SRT.D.11 (apply them to find unknown measurements in right and non-right triangles), together with G-SRT.D.9 (the area formula ½ab sin C). The ambiguous SSA case is the place students most often go wrong: encourage the habit of comparing the opposite side to h = b sin A and sketching the swing arc before trusting a single inverse sine. The real-world measurement problems connect back to right-triangle trig (Stage 25.1) and the circle theorems of Stage 18.