Trigonometry · Trigonometry: Similar Triangles, Right Triangles and Acute TrianglesThe Sine Law
3.1 3.3About 75 minutes
Learning goals- I can follow how the sine law is developed from two right triangles inside an acute triangle.
- I can verify the sine law with measurements.
- I can use the sine law to find an unknown side when I know two angles and a side.
- I can use the sine law to find an unknown angle in an acute triangle when I know two sides and an angle opposite one of them.
Words to know
acute triangle
A triangle whose three angles are all less than 90°.
oblique triangle
Any triangle that is not a right triangle.
altitude
A segment from a vertex perpendicular to the opposite side (or its extension); its length is the height.
sine law
In any triangle ABC, a/sin A = b/sin B = c/sin C (equivalently sin A/a = sin B/b = sin C/c).
opposite pair
A side together with the angle across from it, such as a and ∠A.
Beyond right triangles
SOH CAH TOA only works in right triangles. But most triangles in the real world are not right triangles: two fire towers and a fire, two docks and a buoy, the three corners of a park. In this lesson and the next, you will learn two laws that work in acute triangles (all angles less than 90°): the sine law and the cosine law. (In Grade 11 you will extend them to obtuse triangles.)
Where the sine law comes from
Take an acute triangle ABC with standard labelling. Draw the altitude h from C, perpendicular to AB at D. Now there are two right triangles, and h is a side of both.
- In right △ACD, h is opposite ∠A and b is the hypotenuse: sin A = h/b, so h = b sin A.
- In right △BCD, h is opposite ∠B and a is the hypotenuse: sin B = h/a, so h = a sin B.
These are two equations for the same h, a small system of equations. Setting them equal gives b sin A = a sin B. Dividing both sides by sin A sin B:
a/sin A = b/sin B
Drawing the altitude from A instead gives b/sin B = c/sin C the same way. So:
The sine law
In any acute △ABC:
a/sin A = b/sin B = c/sin C
or, flipped (handy when finding an angle):
sin A/a = sin B/b = sin C/c
You only ever use two of the three fractions at a time.
Check it with measurements
A triangle was drawn with angles 50°, 60° and 70°. Its sides measured a = 8.2 cm, b = 9.2 cm and c = 10.0 cm. Then:
- 8.2 ÷ sin 50° ≈ 10.70
- 9.2 ÷ sin 60° ≈ 10.62
- 10.0 ÷ sin 70° ≈ 10.64
The three ratios agree to within measuring error. In dynamic geometry software (such as GeoGebra or Desmos geometry), you can drag the vertices of a triangle and watch the three ratios stay equal to each other.
When can you use the sine law?
You need one complete opposite pair (a side and the angle across from it) plus one more piece:
- Two angles and a side (AAS or ASA): find the third angle with the angle sum (180°), then you always have a pair. Use the sine law to find a side.
- Two sides and an angle opposite one of them (SSA): use the sine law to find an angle.
If you know two sides and the angle between them (SAS), or all three sides (SSS), there is no complete pair; you will need the cosine law (next lesson).
Worked exampleExample 1: Finding a side (the curriculum's sample problem)
In △ABC, ∠A = 35°, ∠B = 65° and AC = 18 cm. Determine BC.
- Label: AC is opposite ∠B, so b = 18. BC is opposite ∠A, so BC = a.
- Complete pair: b and ∠B. Use a/sin A = b/sin B.
- a/sin 35° = 18/sin 65°.
- Multiply both sides by sin 35°: a = 18 sin 35° ÷ sin 65° = 11.39...
- Check: ∠A (35°) is smaller than ∠B (65°), so a should be shorter than b. 11.4 < 18. ✓
Example 1 drawn to scale. Side a is opposite the 35° angle, and b = 18 cm is opposite the 65° angle.Worked exampleExample 2: Finding an angle
In acute △ABC, a = 13 cm, b = 11 cm and ∠A = 58°. Find ∠B and ∠C to the nearest degree.
- Complete pair: a and ∠A. Unknown angle: ∠B, opposite b. Use the flipped form: sin B/b = sin A/a.
- sin B/11 = sin 58°/13, so sin B = 11 sin 58° ÷ 13 = 0.7175...
- ∠B = sin⁻¹(0.7175...) ≈ 45.9° ≈ 46°.
- ∠C = 180° − 58° − 45.9° ≈ 76°. All three angles are acute. ✓
- Check: b (11) is shorter than a (13), so ∠B should be smaller than ∠A. 46° < 58°. ✓
Worked exampleExample 3: Two angles and the side between them
In △ABC, ∠A = 58°, ∠B = 72° and c = 15 m. Find a and b.
- There is no pair yet. Use the angle sum: ∠C = 180° − 58° − 72° = 50°. Now c and ∠C are a pair.
- a = 15 sin 58° ÷ sin 50° = 16.60... ≈ 16.6 m.
- b = 15 sin 72° ÷ sin 50° = 18.62... ≈ 18.6 m.
- Check: largest angle (72°) is opposite the longest side (b). ✓
A note about angles
In this course every triangle in a sine-law problem is acute, so sin⁻¹ gives the angle you want. In Grade 11 you will meet obtuse triangles, where the same sine can belong to two different angles (the "ambiguous case").
Just for funThe sine law is sometimes called the "law of sines". Either way, it is one of the few laws where nobody minds if you break the triangle into pieces to understand it.
Hands-on activityVerify the sine law on paper
You need: a ruler, a protractor, a sharp pencil, a calculator and plain paper.
- Draw a large acute triangle (sides between 10 cm and 20 cm). Label its vertices A, B and C, and its sides a, b and c in standard position.
- Measure all three sides to the nearest millimetre and all three angles to the nearest degree. Check that the angles add to about 180°.
- Calculate a ÷ sin A, b ÷ sin B and c ÷ sin C.
- Draw a second, differently shaped acute triangle and repeat.
What to look for: within one triangle the three ratios should agree within about 2% (for example, 21.3, 21.6 and 21.4). The ratio is different for different triangles: it depends on the triangle's size. (Fun fact you can test: it equals the diameter of the circle that passes through all three vertices.)
Your turn
1In △ABC, ∠A = 40°, ∠B = 75° and a = 12 cm. Find b.
b/sin 75° = 12/sin 40°, so b = 12 sin 75° ÷ sin 40° ≈ 18.0 cm.
2In △ABC, ∠B = 52°, ∠C = 68° and c = 20 m. Find b.
b = 20 sin 52° ÷ sin 68° ≈ 17.0 m.
3In acute △ABC, ∠A = 64°, a = 15 cm and b = 12 cm. Find ∠B.
sin B = 12 sin 64° ÷ 15 ≈ 0.7190, so ∠B ≈ 46°.
4In acute △ABC, ∠C = 71°, c = 9.5 cm and a = 8.4 cm. Find ∠A.
sin A = 8.4 sin 71° ÷ 9.5 ≈ 0.8360, so ∠A ≈ 57°.
5In △PQR, ∠P = 48°, ∠Q = 62° and PQ = 25 m. Find QR.
∠R = 70°. PQ = r = 25 is opposite ∠R; QR = p is opposite ∠P. p = 25 sin 48° ÷ sin 70° ≈ 19.8 m.
6Write the sine law for △XYZ (sides x, y, z opposite ∠X, ∠Y, ∠Z).
x/sin X = y/sin Y = z/sin Z
7You know b, c and ∠A (the angle between them). Can you use the sine law right away? Explain.
No. You have no complete opposite pair (a is unknown, and ∠B and ∠C are unknown). This is SAS: use the cosine law.
More practice: warm-up
1Write the sine law for △ABC in both forms.
a/sin A = b/sin B = c/sin C and sin A/a = sin B/b = sin C/c.
2∠A = 50° and ∠B = 60°. Find ∠C.
3Solve x/sin 30° = 10/sin 90°.
4sin B = 0.5 and ∠B is acute. Find ∠B.
5In the sine law, each side is paired with which angle?
More practice: on your own
1Solve △ABC: ∠A = 55°, ∠B = 65° and c = 12 cm.
∠C = 60°. a = 12 sin 55° ÷ sin 60° ≈ 11.4 cm. b = 12 sin 65° ÷ sin 60° ≈ 12.6 cm.
2Solve acute △ABC: ∠B = 64°, b = 19 cm and c = 16 cm.
sin C = 16 sin 64° ÷ 19 ≈ 0.7569, so ∠C ≈ 49.2°. ∠A ≈ 180° − 64° − 49.2° = 66.8°. a = 19 sin 66.81° ÷ sin 64° ≈ 19.4 cm.
3Jake wrote a/sin B = b/sin A. What is wrong?
Each side must be over the sine of its opposite angle. Correct: a/sin A = b/sin B.
4A triangle has angles 40°, 60° and 80°. The side opposite 40° is 6.4 cm. Find the side opposite 80°.
x = 6.4 sin 80° ÷ sin 40° ≈ 9.8 cm.
5Explain, using the altitude h from C, why b sin A = a sin B.
In right △ACD, h = b sin A. In right △BCD, h = a sin B. Both expressions equal the same h, so b sin A = a sin B.
Challenge
1Use the altitude from A (perpendicular to BC) to show that b/sin B = c/sin C.
Call the altitude k. In the right triangle containing ∠B, k is opposite ∠B and c is the hypotenuse: k = c sin B. In the one containing ∠C, k = b sin C. So c sin B = b sin C, and dividing by sin B sin C gives c/sin C = b/sin B.
2Two surveyors stand at A and B on one bank of a river, 300 m apart. They sight a tree C on the far bank: ∠CAB = 70° and ∠CBA = 65°. Find AC, and then the width of the river (the perpendicular distance from C to line AB).
∠C = 45°. AC = b = 300 sin 65° ÷ sin 45° ≈ 384.5 m. Width = AC sin 70° ≈ 361.3 m (right-triangle trigonometry with the altitude from C).
3In an acute triangle, explain why the longest side is always opposite the largest angle.
By the sine law, each side equals the same constant × the sine of its opposite angle. For angles between 0° and 90°, a larger angle has a larger sine, so the largest angle gives the longest side.
Quiz
1. Which is the sine law for △ABC?
2. In △ABC, ∠A = 35°, ∠B = 65° and AC = 18 cm. What is BC?
3. In which case can you NOT start with the sine law?
4. In the development of the sine law, the altitude h from C is written two ways. Which?
5. ∠A = 58° and ∠B = 72°. What is ∠C?
6. In acute △ABC, a = 13, b = 11 and ∠A = 58°. What is ∠B (nearest degree)?
Big idea: Expectations 3.1 and 3.3: exploring how the sine law is developed (two right triangles sharing an altitude, which is a small system of equations) and using it to find sides and angles of acute triangles. Students are not required to reproduce the development on their own, but following it helps them remember the law.
How to help: Ask your teen to circle the complete pair (a side and its opposite angle) before writing anything. If there is no pair and two angles are known, the angle sum gives one.
Common difficulties:
- Pairing a side with an angle that is not opposite it.
- Forgetting to find the third angle first (Example 3).
- Using the sine law in an SAS or SSS situation, where it cannot start.
- Calculator in radian mode.