Grade by Grade
C. Trigonometric Functions · Trigonometry of Acute Triangles

Using the Sine Law to Find Angles and Solve Triangles

C1.4About 70 minutes
Learning goals
  • I can use the sine law to find an unknown angle when I know two sides and the angle opposite one of them.
  • I can solve an acute triangle completely using the sine law and the angle sum.
  • I can describe when the sine law can be used: I need a complete opposite pair plus one more side or angle.
  • I can check that my answers make sense (largest angle across from longest side, angles add to 180°).

Words to know

SSA
Two sides and an angle that is NOT between them (the angle is opposite one of the sides).
AAS / ASA
Two angles and a side. In ASA the side is between the angles; in AAS it is not.
solve a triangle
Find all three sides and all three angles.

Running the sine law backwards

Last lesson you used the sine law to find sides. It can also find angles. This matters in real problems: surveyors often measure two distances and one angle, then need the other angles to finish a map.

When the unknown is an angle, flip every fraction so the sines are on top:

sin A/a = sin B/b = sin C/c

Then solve for the sine and use sin⁻¹.

Finding a SIDEa / sin A = b / sin Bunknown side on topFinding an ANGLEsin A / a = sin B / bunknown sine on top, then sin⁻¹
Two forms of the same law. Use the form that puts your unknown in the numerator.
Worked example

Example 1: Finding an angle (SSA)

In acute △ABC, a = 17.5 cm, b = 14.2 cm and ∠A = 62°. Find ∠B and ∠C.

  1. Complete pair: a = 17.5 and ∠A = 62°. The unknown angle ∠B is across from b = 14.2.
  2. sin B / 14.2 = sin 62° / 17.5
  3. sin B = 14.2 sin 62° ÷ 17.5 ≈ 0.7164
  4. ∠B = sin⁻¹(0.7164) ≈ 45.8°
  5. ∠C = 180° − 62° − 45.8° ≈ 72.2°
  6. Check: b < a, so ∠B should be less than ∠A. It is. ✓

∠B ≈ 46°, ∠C ≈ 72°

Why "acute" matters

Your calculator's sin⁻¹ always gives an angle between 0° and 90°. In this course every triangle is acute, so that is the angle you want. In other courses (such as Functions, MCR3U) you will learn that a sine value can also belong to an obtuse angle, which can create two possible triangles. If a question here says the triangle is acute, you can trust sin⁻¹.

Worked example

Example 2: Solve the triangle

Solve acute △ABC given ∠C = 58°, c = 24 cm and b = 21 cm.

  1. Pair: c = 24 and ∠C = 58°. Find ∠B first (across from the known side b).
  2. sin B = 21 sin 58° ÷ 24 ≈ 0.7420, so ∠B ≈ 47.9°.
  3. ∠A = 180° − 58° − 47.9° ≈ 74.1°.
  4. a / sin 74.1° = 24 / sin 58°, so a ≈ 24 sin 74.1° ÷ sin 58° ≈ 27.2 cm. (Use the unrounded ∠A in your calculator.)
  5. Check: largest angle ∠A ≈ 74° is across from the longest side a ≈ 27.2 cm. ✓

∠B ≈ 48°, ∠A ≈ 74°, a ≈ 27.2 cm

58°ABCc = 24 cmb = 21 cma = ??Pair: c = 24 and ∠C = 58°sin B = 21 sin 58° ÷ 24∠B ≈ 48°∠A ≈ 180° − 58° − 48° ≈ 74°a ≈ 27.2 cm
Example 2, to scale. The known pair (c and ∠C) unlocks ∠B, then the angle sum gives ∠A, then the sine law gives a.
Worked example

Example 3: A plot of land

A triangular plot XYZ has XZ = 120 m and YZ = 105 m. The corner angle at Y is 72°. The plot is acute. Find the other two corner angles and the length of the third side XY.

  1. Pair: XZ = 120 m is across from ∠Y = 72°.
  2. YZ = 105 m is across from ∠X: sin X = 105 sin 72° ÷ 120 ≈ 0.8322, so ∠X ≈ 56.3°.
  3. ∠Z = 180° − 72° − 56.3° ≈ 51.7°.
  4. XY is across from ∠Z: XY = 120 sin 51.7° ÷ sin 72° ≈ 99.0 m.

∠X ≈ 56°, ∠Z ≈ 52°, XY ≈ 99 m

72°XYZ120 m105 mXY = ?
Example 3, to scale. The 120 m side faces the 72° corner, which gives the complete pair.

When can you use the sine law?

You need one complete opposite pair plus one more piece:

  • AAS or ASA (two angles and any side): find the third angle if needed, then find sides.
  • SSA (two sides and an angle opposite one of them): find the angle opposite the other known side.

If you know two sides and the angle between them (SAS) or all three sides (SSS), there is no complete pair. Those cases need the cosine law, coming in the next two lessons.

Just for fun

Riddle: why did the sine law refuse to work with SAS?

It said, "Sorry, I only work with pairs. Come back when you've found your other half."

Hands-on activity

Solve a triangle on the floor

You need: three pieces of string or tape, a measuring tape, a large protractor (or a paper one you make by tracing), and some floor space.

  1. Tape a large acute triangle on the floor (sides 1 m to 2 m).
  2. Measure two sides and the angle across from one of them. Do not measure anything else yet.
  3. Use the sine law to calculate the other two angles and the third side.
  4. Now measure them. How close were you?

What to look for: angles within 2° to 3° and the side within a few centimetres. If one value is far off, check whether you used the angle opposite the side.

Your turn

1

In acute △ABC, a = 12 cm, b = 10 cm and ∠A = 70°. Find ∠B.

2

In acute △PQR, q = 9.4 m, r = 8.1 m and ∠Q = 66°. Find ∠R and ∠P.

3

Solve acute △ABC: ∠B = 54°, b = 30 cm, c = 28 cm.

4

Solve △DEF: ∠D = 44°, ∠E = 71°, d = 15 m.

5

For each set of givens, say whether the sine law can be used right away: (a) a, b, ∠A (b) a, b, ∠C (c) ∠A, ∠B, c (d) a, b, c.

6

A student found sin B = 1.18. What does this tell you?

More practice: warm-up

1

Find ∠B if sin B = 0.7071 and ∠B is acute.

2

Solve sin B / 8 = 0.1.

3

Write the sine law with sines on top.

4

In △ABC, ∠A = 72° and ∠B = 51°. Find ∠C.

5

Which known pieces make up SSA?

More practice: on your own

1

Two sides of an acute triangular garden are 6.5 m and 5.8 m. The angle across from the 6.5 m side is 74°. Find the angle across from the 5.8 m side.

2

Solve acute △XYZ: x = 41 cm, z = 36 cm, ∠X = 77°.

3

Lena wrote sin B = 14.2 × 17.5 ÷ sin 62° for Example 1. Explain the error.

4

A triangular sail has edges 4.8 m and 4.2 m. The angle across from the 4.8 m edge is 68°. The sail is acute. Find all its angles and the third edge.

5

In acute △ABC, a = 9.0, ∠A = 50° and ∠B = 60°. Find b, then use b and ∠B (instead of a and ∠A) to find c. Compare with using a and ∠A.

Challenge

1

In △ABC, a = 8, b = 12 and ∠A = 50°. Show that no triangle exists.

2

In an acute triangle, ∠A = 40° and a/sin A = 20. Find a, and find the largest possible length any side of this triangle could have.

3

Prove that if ∠A = ∠B in a triangle, then a = b, using the sine law.

Check your understanding

Quiz

1.

Which form of the sine law is easiest for finding an unknown angle B?

2.

In acute △ABC, a = 17.5, b = 14.2, ∠A = 62°. What is ∠B?

3.

Which set of givens can NOT be started with the sine law?

4.

Why can you trust sin⁻¹ in this unit?

5.

A calculation gives sin C = 1.06. What does that mean?

6.

After solving a triangle, which check is useful?

For parents and teachers

Big idea: Expectation C1.4: describe the conditions that guide when the sine law is appropriate, and use it to calculate sides and angles in acute triangles.

How to help: Ask your teen: "Which pair is complete? What is across from the angle you want?" Then have them check that the angles add to 180°.

Common difficulties:

  • Rearranging sin B/b = sin A/a incorrectly (multiplying by the wrong side).
  • Forgetting sin⁻¹ at the end and reporting a sine (like 0.74) as the angle.
  • Using a rounded angle in the next step.
  • Trying the sine law on SAS or SSS (it cannot start).