The sine rule — finding sides and angles — Mathematics

Before this

This leads to

A triangle with no right angle

Two lookouts stand on a straight stretch of coast, 500 m apart. Call their positions A and B. Both can see the same boat out at sea. The lookout at A measures the angle between the coast and the boat: 62°. The lookout at B measures 48°. How far is the boat from A?

Look at the triangle the lookouts and the boat make. It has no right angle. That is a real problem, because every tool you have so far needs one: SOH CAH TOA and Pythagoras both work only in a right-angled triangle.

You could make a right angle. Drop a line from the boat to the coast, at 90°. Now there are two right-angled triangles. But you know neither where that line meets the coast nor how long it is.

Most real triangles have no right angle. This lecture builds one rule for all of them. And the idea that cracks it is the line you just drew, even though you never find its length.

Names first: each side faces an angle

We need names that work for every triangle. Call the corners A, B and C, and use the same letters for the angles there. Then each side gets the small letter of the angle across from it. Side $a$ is across from angle A: it is the side that does not touch A. In the same way, $b$ faces B and $c$ faces C.

So a side and the angle facing it share a letter. We call them a pair: $a$ with A, $b$ with B, $c$ with C. This is the same idea as "opposite" in SOH CAH TOA. The difference is that now every angle has its own opposite side, not just one chosen angle.

Each side is named by the angle across from it. The dashed height h from C to the base splits the triangle into two right-angled triangles, ACH and BCH.

Where the rule comes from: one height, written two ways

Now use the line from the hook: the height $h$ from C to H on AB. You do not know $h$ , and you do not need to. You only need to write it in two different ways. Step through the argument below.

Step through — the sine rule from one height

Press Next step to add one move. Each move comes with the reason it is allowed.

Each step below gives one line of the argument and the reason it is allowed. The triangle is the one in the figure above.

Step 1. Draw the height $h$ from C to H on AB. Reason: a perpendicular makes two right-angled triangles, ACH and BCH.
Step 2. In triangle ACH, $sin A = \frac{h}{b}$ . Reason: $h$ is opposite angle A, and $b$ is the hypotenuse.
Step 3. So $h = b sin A$ . Reason: multiply both sides by $b$ .
Step 4. In triangle BCH, $sin B = \frac{h}{a}$ , so $h = a sin B$ . Reason: the same height, now seen from B, where $a$ is the hypotenuse.
Step 5. $a sin B = b sin A$ . Reason: both expressions equal the same length $h$ .
Step 6. $\frac{a}{sin A} = \frac{b}{sin B}$ . Reason: divide both sides by $sin A sin B$ , which is not zero.
Step 7. A height from A to side BC gives $\frac{b}{sin B} = \frac{c}{sin C}$ in the same way. Reason: the argument works from any corner.

Conclusion. All three fractions are equal: a/sin A = b/sin B = c/sin C.

The height was unknown at the start and at the end. It was only a bridge that linked each side to the angle facing it.

What if one angle is obtuse?

The argument above has a hidden assumption. It assumed that H lands between A and B. In a triangle with an obtuse angle at B, that is not true. The height from C now lands outside the triangle, on the base line extended beyond B.

When angle B is obtuse, the height from C lands outside the triangle. The right-angled triangle CBH has the angle 180° − B at B, not B itself.

Triangle ACH still gives $h = b sin A$ . But the right-angled triangle on the other side is now CBH, and its angle at B is not B. Angles on a straight line add to 180°, so it is $18 0^{\circ} - B$ . That triangle gives $h = a sin (18 0^{\circ} - B)$ .

So the proof works only if $sin B$ means the same number as $sin (18 0^{\circ} - B)$ . That is exactly how the sine of an obtuse angle is defined, and it is what your calculator does. Try it: $sin 15 0^{\circ} = 0.5$ , the same as $sin 3 0^{\circ}$ . The lecture on missing angles promised that this fact would come back in triangles with an obtuse angle. Here it is. It is the reason the sine rule works in every triangle.

One fact to carry with you

For any angle $θ$ between 0° and 180°, $sin θ = sin (18 0^{\circ} - θ)$ . An angle and its partner that makes 180° have the same sine. The unit circle at Pre-University shows why this definition is the natural one.

The sine rule

In any triangle ABC, with side $a$ opposite angle A, side $b$ opposite angle B and side $c$ opposite angle C:

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}

To find an angle, it is easier to use the same rule upside down:

\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}

Choose the form that puts your unknown on top. In words: every side divided by the sine of the angle facing it gives the same number. That is it. That is the whole idea. The rest of this lecture is about using it well.

Stop and think

Does the sine rule still work in a right-angled triangle? Try it with C = 90°.

Show the answer

Yes. Since $sin 9 0^{\circ} = 1$ , the rule says $\frac{a}{sin A} = \frac{c}{1}$ . So $a = c sin A$ , or $sin A = \frac{a}{c}$ . That is "sine = opposite ÷ hypotenuse", because $c$ faces the right angle. The sine rule is not a new idea that replaces SOH CAH TOA. It is SOH CAH TOA, extended to every triangle.

When can you use it? You need one complete pair

To use the rule, you must know both parts of one pair, plus one more fact: a second angle or a second side. The unknown is the other half of that second pair.

Two angles and a side. If you know two angles, you know the third, because the angles add to 180°. So you always have a complete pair. Books call these cases AAS and ASA.
Two sides and an angle that is not between them. The angle faces one of the known sides, so that is your pair. Books call this SSA. It has a surprise, which the second half of this lecture explains.

Two cases do not give a complete pair: three sides (SSS), and two sides with the angle between them (SAS). For those, the next lecture builds the cosine rule.

Worked examples: finding a side

Worked Example 1 — Two angles and the side facing one of them

In triangle ABC, A = 35°, B = 72° and $a = 8$ cm. Find $b$ .

Think first. Which pair is complete? Side $a$ and angle A. The unknown, $b$ , pairs with B, which we know.

\frac{b}{sin 7 2 ^{\circ}} = \frac{8}{sin 3 5 ^{\circ}}

The unknown is on top, so multiply both sides by $sin 7 2^{\circ}$ :

b = \frac{8 sin 7 2 ^{\circ}}{sin 3 5 ^{\circ}}

b = \frac{8 \times 0.9511}{0.5736} \approx 13.3 cm

Type it into the calculator in one go; the decimals only show the size of each part.

Check. The larger angle faces the longer side. B = 72° is larger than A = 35°, so $b$ must be longer than 8 cm. It is. ✓

Worked Example 2 — The boat from the hook

The lookouts are 500 m apart, with angles 62° at A and 48° at B. Call the boat C. Find the distance AC.

Think first. The side we know, AB = 500 m, faces the angle at the boat, C. We do not know C yet. So the first move is to find it.

C = 18 0^{\circ} - 6 2^{\circ} - 4 8^{\circ} = 7 0^{\circ}

Now the pair ( $c$ , C) = (500 m, 70°) is complete. The distance AC is side $b$ , because it faces the angle at B, 48°.

b = \frac{500 sin 4 8 ^{\circ}}{sin 7 0 ^{\circ}} \approx 395 m

Check. 48° is the smallest angle, so $b$ should be the shortest side. Side $a = 500 sin 6 2^{\circ} \div sin 7 0^{\circ} \approx 470$ m and $c = 500$ m, so it is. ✓

Worked Example 3 — An exact answer

In triangle PQR, P = 30°, Q = 45° and $p = 6$ cm. Find $q$ exactly.

Think first. Examiners choose 30° and 45° so that you can use the exact values instead of a calculator. From the half-square and the half-equilateral triangle, $sin 3 0^{\circ} = \frac{1}{2}$ and $sin 4 5^{\circ} = \frac{2}{2}$ .

q = \frac{6 sin 4 5 ^{\circ}}{sin 3 0 ^{\circ}}

q = \frac{6 \times \frac{2}{2}}{\frac{1}{2}} = 62 cm

Dividing by $\frac{1}{2}$ is the same as multiplying by 2, and the 2s cancel. So $q = 62 \approx 8.49$ cm.

Check. Q is larger than P, so $q$ should be longer than 6 cm. ✓

Finding an angle, and the second answer

To find an angle, use the upside-down form, so that the sine you want is on top. Then use $sin^{- 1}$ , as in the lecture on missing angles. But there is one new step. The calculator gives one angle. Since $sin θ = sin (18 0^{\circ} - θ)$ , there is always a second angle with the same sine. You must decide whether that second angle can also fit in the triangle.

Worked Example 4 — Only one angle fits

In triangle ABC, $a = 10$ cm, $b = 7$ cm and A = 50°. Find B.

Think first. The pair ( $a$ , A) is complete, and B pairs with $b = 7$ .

\frac{sin B}{7} = \frac{sin 5 0 ^{\circ}}{10}

sin B = \frac{7 sin 5 0 ^{\circ}}{10} \approx 0.5362

The calculator gives $B = sin^{- 1} (0.5362 \dots) \approx 32. 4^{\circ}$ . The second angle with the same sine is $18 0^{\circ} - 32. 4^{\circ} = 147. 6^{\circ}$ .

Can B be 147.6°? Then A + B = 50° + 147.6° = 197.6°. That is more than 180°, so there is no room for angle C. The second answer is impossible. So B ≈ 32.4°, and C ≈ 180° − 50° − 32.4° = 97.6°.

Check. Side $b = 7$ is shorter than $a = 10$ , so B must be smaller than A. It is: 32.4° is less than 50°. ✓

Notice why the second answer failed. Side $b$ was shorter than $a$ , so B had to be smaller than A, and an obtuse B could not be. When the side facing the known angle is the longer one, there is only one answer.

The ambiguous case: one set of facts, two triangles

Now change the numbers so that the side facing the known angle is the shorter one. In triangle ABC, $a = 9$ cm, $b = 12$ cm and A = 40°. Find B.

sin B = \frac{12 sin 4 0 ^{\circ}}{9} \approx 0.8571

The calculator gives B ≈ 59.0°. The second angle is 180° − 59.0° = 121.0°. Test it: 40° + 121.0° = 161.0°, which leaves 19.0° for C. That is a real triangle. So both answers fit.

This is not a mistake in the algebra. It is a fact about triangles. Picture it. Angle A and side $b = 12$ cm fix the corner C. Side $a = 9$ cm must reach from C down to the base line. Imagine it as a stick hinged at C. It can swing to meet the base line at two different points. Each point gives a triangle, and both have A = 40°, $b = 12$ and $a = 9$ .

Move the length of side $a$ below and watch how many triangles fit.

Explore — how many triangles fit?

Angle A is 40° and side b is 12 cm. Change side a. The dotted circle shows every place side a can reach from C. Count where it meets the base line.

Side a: 9.0 cm

This box shows how many triangles have A = 40°, b = 12 cm and a given side a. The height from C to the base line is 12 sin 40° ≈ 7.71 cm.

With a = 9 cm, side a can meet the base line at B₁ or at B₂. Both triangles have A = 40°, b = 12 cm and a = 9 cm.

Side a (cm)	sin B	Triangles	Angle B
6	≈ 1.286	none	—
≈ 7.71	1	one	90°
9	≈ 0.857	two	≈ 59.0° or 121.0°
14	≈ 0.551	one	≈ 33.4°

At 6 cm, sin B would be above 1, which is impossible. At 14 cm, the second angle, 146.6°, leaves no room for C.

You should have seen four situations, all decided by the height from C to the base, $12 sin 4 0^{\circ} \approx 7.71$ cm.

Shorter than the height: side $a$ cannot reach the base. No triangle, and the rule gives $sin B$ above 1.
Equal to the height: it just touches the base. One triangle, with B = 90°.
Between the height and $b$ : it meets the base twice. Two triangles.
At least as long as $b$ : the second point falls behind A. One triangle.

This is the ambiguous case: the three facts do not decide the triangle. It needs SSA with an acute given angle, and the side facing it the shorter of the two. If the given angle is obtuse, the other two angles must both be acute, so there is one triangle or none.

In an exam, you do not need circles. Find the calculator's angle and 180° minus it, and check whether each leaves room for the third angle.

Three traps

Trap 1 — A side paired with the wrong angle

Each fraction is a side over the sine of the angle facing it, never an angle at one end of the side. Put a finger on the angle and go straight across the triangle. The side you reach is its partner.

Trap 2 — Trusting the calculator's single angle

$sin^{- 1}$ gives one answer between 0° and 90°. The second angle, 180° minus the first, has the same sine. Test it every time, and say why it fits or fails.

Trap 3 — No complete pair

Given $a = 7$ , $b = 9$ and C = 50°, a student writes $\frac{7}{sin A} = \frac{9}{sin B}$ . This equation has two unknowns, so it cannot be solved. The known angle sits between the two known sides, so it faces neither of them. This is SAS, and it needs the cosine rule.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

In triangle XYZ, X = 28°, Y = 104° and $x = 6.5$ cm. Find $y$ and $z$ , to 3 significant figures.

Hint 1
Which pair is complete? You will also need the third angle for $z$ .

Hint 2
Z = 180° − 28° − 104°. Then $y = \frac{6.5 sin 10 4 ^{\circ}}{sin 2 8 ^{\circ}}$ .

Full answer
The pair ( $x$ , X) = (6.5 cm, 28°) is complete. $y = \frac{6.5 sin 10 4 ^{\circ}}{sin 2 8 ^{\circ}} \approx 13.4$ cm. Z = 180° − 28° − 104° = 48°, so $z = \frac{6.5 sin 4 8 ^{\circ}}{sin 2 8 ^{\circ}} \approx 10.3$ cm. Check: the angles in order are 28° < 48° < 104°, and the sides in the same order are 6.5 < 10.3 < 13.4. ✓ Y is obtuse, and the calculator uses $sin 10 4^{\circ} = sin 7 6^{\circ}$ without any extra step. ✓
In triangle ABC, $a = 11$ cm, $b = 8$ cm and A = 63°. Find B and C. Explain why there is only one answer.

Hint 1
Use the upside-down form: $sin B = \frac{8 sin 6 3 ^{\circ}}{11}$ .

Hint 2
Find the second angle, 180° minus the calculator's angle. Add it to 63°.

Full answer
$sin B = \frac{8 sin 6 3 ^{\circ}}{11} \approx 0.6480$ , so B ≈ 40.4°. The second angle is about 139.6°, but 63° + 139.6° = 202.6°, which is more than 180°. So B ≈ 40.4° only, and C ≈ 180° − 63° − 40.4° = 76.6°. The reason in one line: $b$ is shorter than $a$ , so B must be smaller than A, and an obtuse B cannot be. Check: $c = \frac{11 sin 76. 6 ^{\circ}}{sin 6 3 ^{\circ}} \approx 12.0$ cm is the longest side, and it faces the largest angle. ✓
Angle A = 30° and side $b = 10$ cm are fixed. For each length of side $a$ , say how many triangles ABC are possible: (a) 4 cm, (b) 5 cm, (c) 7 cm, (d) 12 cm. Give the angle or angles B where they exist.

Hint 1
Work out $sin B = \frac{10 sin 3 0 ^{\circ}}{a} = \frac{5}{a}$ in each case.

Hint 2
The height from C to the base is $10 sin 3 0^{\circ} = 5$ cm. Compare each $a$ with 5 and with 10.

Full answer
(a) $sin B = 1.25$ . No angle has a sine above 1, so there is no triangle: 4 cm is shorter than the height of 5 cm. (b) $sin B = 1$ , so B = 90°: one right-angled triangle. (c) $sin B = \frac{5}{7} \approx 0.7143$ , so B ≈ 45.6° or B ≈ 134.4°. Both fit (30° + 134.4° = 164.4°), so there are two triangles, with C ≈ 104.4° or C ≈ 15.6°. (d) $sin B = \frac{5}{12} \approx 0.4167$ , so B ≈ 24.6°. The second angle, 155.4°, fails: 30° + 155.4° = 185.4°. One triangle. Check: this matches the rule from the explorer. Below the height, none; equal, one; between 5 and 10, two; 10 or more, one. ✓
A stretch. A surveyor stands at P on level ground and measures the angle of elevation of the top T of a tower: 25°. She walks 100 m straight towards the tower, to Q, and measures 40°. Use the sine rule in triangle PQT to find the height of the tower.

Hint 1
Sketch triangle PQT. The angle at Q inside it is 180° − 40° = 140°. So the angle at T is 180° − 25° − 140°.

Hint 2
The angle at T is 15°, and it faces PQ = 100 m. Find QT with the sine rule, then use the right-angled triangle at the foot of the tower.

Full answer
In triangle PQT, the angle at Q is 180° − 40° = 140° (angles on a straight line), so the angle at T is 180° − 25° − 140° = 15°. The pair (PQ, T) = (100 m, 15°) is complete. QT faces the 25° angle at P, so $QT = \frac{100 sin 2 5 ^{\circ}}{sin 1 5 ^{\circ}} \approx 163.3$ m. Now QT is the hypotenuse of the right-angled triangle from Q to the foot of the tower, with the angle of elevation 40° at Q. So the height is $QT sin 4 0^{\circ} \approx 105$ m (104.96 m). Check by a second route, with tan: if the tower is $h$ high, then $\frac{h}{tan 2 5 ^{\circ}} - \frac{h}{tan 4 0 ^{\circ}} = 100$ , which also gives $h \approx 104.96$ m. ✓

From a 12 km line to a map of a country

The sine rule is how countries were mapped before satellites. The method is called triangulation. You measure one length on the ground very carefully: the baseline. From its two ends, you measure the angles to a far-off hilltop. Now you know two angles and the side between them, so the sine rule gives the other two sides. Each new side becomes the known side of the next triangle. So a chain of triangles can cross a country from one measured length.

The Great Trigonometrical Survey of India began this way. On 10 April 1802, William Lambton's team started measuring a baseline near Madras (now Chennai), from St Thomas Mount to Perumbauk Hill. It was about 7.5 miles, or 12.1 km, long, and took until 22 May to measure. In 1804 a second baseline was measured near Bangalore, as a test. Its length, calculated through the chain of triangles from Madras, differed from the measured length by only about 3.5 inches, or 8.9 cm. (Sources: Wikipedia, "Great Trigonometrical Survey" and "William Lambton".)

The bigger idea

The sine rule says that a side divided by the sine of its angle is the same number for all three pairs. What is that number? It has a meaning, and the circle theorems reveal it.

Every triangle fits inside exactly one circle through its three corners. Call its centre O and its radius R. Draw the diameter from B through O, and call its other end D. The angle BCD is 90°, because it is the angle in a semicircle. The angle at D stands on the same arc BC as the angle at A, so it equals A. Now look at the right-angled triangle BCD. Its hypotenuse is the diameter, 2R, and the side opposite D is $a$ . So $sin A = \frac{a}{2 R}$ .

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2 R

The common number is the diameter of the circle through the corners. In Worked Example 1, it is $8 \div sin 3 5^{\circ} \approx 13.9$ cm. If A is obtuse, D lands on the other side of BC. Then the cyclic quadrilateral gives the angle 180° − A, which has the same sine. Trigonometry and circle geometry, which looked like separate topics, are telling the same story.

Key concepts in this lesson

In triangle ABC, side $a$ faces angle A, side $b$ faces angle B and side $c$ faces angle C; a side and its facing angle form a pair.
The sine rule, $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ , comes from writing the same height of the triangle in two ways.
For any angle between 0° and 180°, $sin θ = sin (18 0^{\circ} - θ)$ , which is why the rule also works when one angle is obtuse.
The sine rule needs one complete pair plus one more angle or side, so it solves AAS, ASA and SSA triangles but not SSS or SAS.
When finding an angle, always test the second angle, 180° minus the calculator's answer, to see whether it leaves room for the third angle.
In the ambiguous case, two different triangles fit the same SSA facts: this happens when the side facing the acute given angle is longer than the height but shorter than the other given side.