C. Trigonometric Functions · Trigonometry of Acute TrianglesDiscovering and Verifying the Sine Law
C1.3 C1.4About 75 minutes
Learning goals- I can verify the sine law with measurements and with technology such as a spreadsheet or dynamic geometry software.
- I can explain where the sine law comes from, using an altitude and two right triangles.
- I can label a triangle in standard form and identify opposite side–angle pairs.
- I can use the sine law to find a side when I know two angles and a side.
Words to know
acute triangle
A triangle in which all three angles are less than 90°.
oblique triangle
A triangle with no right angle.
standard labelling
Vertices A, B, C with the side opposite each one named by the same lower-case letter: a is across from ∠A, and so on.
sine law
In any acute triangle ABC, a/sin A = b/sin B = c/sin C. (It also works in every other triangle.)
opposite pair
A side together with the angle directly across from it, such as b and ∠B.
No right angle? No problem.
Two forest-fire lookouts see the same smoke. A surveyor measures a lot whose corners aren't square. A truss has a peak that isn't 90°. None of these triangles has a right angle, so SOH CAH TOA can't be used directly. In this lesson you will discover a pattern that works in every acute triangle, check it with technology, and see why it must be true. It is called the sine law. (If you took MPM2D, you met the sine law there. This time the focus is on verifying it, including with technology, and on deciding when it applies.)
Standard labelling: name the vertices A, B and C. The side across from ∠A is a, the side across from ∠B is b, and the side across from ∠C is c.
Investigate: is there a pattern?
A student drew a large acute triangle, measured its angles with a protractor and its sides with a ruler, and then divided each side by the sine of the opposite angle. Then she did the same for a second, smaller triangle.
| Triangle | Side | Opposite angle | side ÷ sin(angle) |
|---|
| 1 | a = 9.0 cm | ∠A = 40° | 14.00 |
| 1 | b = 12.7 cm | ∠B = 65° | 14.01 |
| 1 | c = 13.5 cm | ∠C = 75° | 13.98 |
| 2 | a = 7.7 cm | ∠A = 50° | 10.05 |
| 2 | b = 8.7 cm | ∠B = 60° | 10.05 |
| 2 | c = 9.4 cm | ∠C = 70° | 10.00 |
Verify with technology
Measurements always have small errors. Technology lets you test many triangles quickly and exactly.
- Dynamic geometry software (such as GeoGebra, free online): draw a triangle, have the program measure the sides and angles, and type a calculation such as a/sin(α). Drag a vertex around: the three ratios change together but stay equal to each other, as long as the triangle exists.
- Spreadsheet: put the measurements in columns and let a formula calculate each ratio. Spreadsheets use radians, so a formula looks like
=B2/SIN(RADIANS(C2)) when B2 holds a side and C2 holds its opposite angle in degrees.
Why it works
Draw the altitude h from C, meeting AB at D. Now there are two right triangles that share h, just like the two-triangle problems from earlier in this unit.
- In △ACD, sin A = h/b, so h = b sin A.
- In △BCD, sin B = h/a, so h = a sin B.
Both equal h, so b sin A = a sin B. Divide both sides by sin A sin B to get a/sin A = b/sin B. Drawing the altitude from A instead gives b/sin B = c/sin C the same way.
The sine law
In any acute △ABC:
a/sin A = b/sin B = c/sin C
or, flipped (handy for finding angles in the next lesson):
sin A/a = sin B/b = sin C/c
You only ever use two of the fractions at a time, and you need one complete opposite pair (a side and its opposite angle, both known).
Worked exampleExample 1: Two angles and a side not between them (AAS)
In acute △ABC, ∠A = 48°, ∠B = 67° and a = 21 m. Find b and c.
- Complete pair: a = 21 and ∠A = 48°.
- b/sin 67° = 21/sin 48°, so b = 21 sin 67° ÷ sin 48° ≈ 26.01 m.
- For c, first find ∠C = 180° − 48° − 67° = 65°.
- c = 21 sin 65° ÷ sin 48° ≈ 25.61 m.
- Check: the largest angle (67°) is across from the longest side (b). ✓
Worked exampleExample 2: Two angles and the side between them (ASA)
In △PQR, ∠P = 52°, ∠Q = 74° and PQ = 30 cm. Find QR and PR.
- There is no complete pair yet. Use the angle sum: ∠R = 180° − 52° − 74° = 54°. PQ is opposite ∠R, so r = 30 and ∠R = 54° make a pair.
- QR = p (opposite ∠P): p = 30 sin 52° ÷ sin 54° ≈ 29.2 cm.
- PR = q (opposite ∠Q): q = 30 sin 74° ÷ sin 54° ≈ 35.6 cm.
QR ≈ 29.2 cm, PR ≈ 35.6 cm
Worked exampleExample 3: A roof truss
The bottom chord of a roof truss is 9.0 m long. The left rafter makes a 50° angle with the chord and the right rafter makes a 62° angle. How long is each rafter?
- The peak angle is 180° − 50° − 62° = 68°, across from the 9.0 m chord. That is the complete pair.
- The left rafter is across from the 62° angle: 9.0 sin 62° ÷ sin 68° ≈ 8.57 m.
- The right rafter is across from the 50° angle: 9.0 sin 50° ÷ sin 68° ≈ 7.44 m.
The left rafter is about 8.57 m and the right rafter is about 7.44 m long.
Example 3, to scale. Each rafter is across from the base angle at the OTHER end of the chord. That is the most common place to mix up pairs.Just for funA funny-but-true fact: the constant a/sin A is the diameter of the circle that passes through all three corners of the triangle. So the sine law is secretly a circle law wearing a triangle costume.
Hands-on activityVerify the sine law yourself
You need: plain paper, a ruler, a protractor, a sharp pencil and a calculator (or a free spreadsheet or GeoGebra on a computer).
- Draw a large acute triangle (sides 10 cm to 20 cm). Label it in standard form.
- 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 differently shaped acute triangle and repeat. If you have a computer, put your numbers in a spreadsheet with
=B2/SIN(RADIANS(C2)). - Bonus (technology): in GeoGebra, build a triangle, show the ratios, and drag a vertex. Write down what changes and what stays the same.
What to look for: the three ratios in one triangle agree within about 2%. The ratios for two different triangles are usually different. Bonus: the three ratios always stay equal to each other while you drag.
Your turn
1Write the sine law for △XYZ.
x/sin X = y/sin Y = z/sin Z
2In △ABC, ∠A = 55°, ∠B = 72° and a = 14 cm. Find b.
b = 14 sin 72° ÷ sin 55° ≈ 16.3 cm
3In △ABC, ∠B = 61°, ∠C = 49° and b = 8.6 m. Find c.
c = 8.6 sin 49° ÷ sin 61° ≈ 7.4 m
4In △DEF, ∠D = 64°, ∠E = 58° and DE = 40 cm. Find EF and DF.
∠F = 58°. DE = f = 40. EF = d = 40 sin 64° ÷ sin 58° ≈ 42.4 cm. DF = e = 40 sin 58° ÷ sin 58° = 40 cm (∠E = ∠F, so the triangle is isosceles).
5Measured values: a = 6.2 cm, ∠A = 44°, b = 8.1 cm, ∠B = 65°. Do they support the sine law?
6.2 ÷ sin 44° ≈ 8.93 and 8.1 ÷ sin 65° ≈ 8.94. These agree to within about 0.1%, well within measuring error, so yes.
6A spreadsheet has a side in cell B3 and its opposite angle (in degrees) in C3. Write a formula for the ratio.
7In Example 1, find c using the pair b and ∠B instead of a and ∠A. Do you get the same answer?
c = b sin 65° ÷ sin 67° with b ≈ 26.012: c ≈ 25.61 m. Same answer, as expected. (Using the given pair is safer: it has no rounding.)
More practice: warm-up
1Which side is opposite ∠C in △ABC?
2Two angles of a triangle are 70° and 45°. Find the third.
3Solve x/sin 30° = 12/sin 90°.
4Evaluate 10 ÷ sin 45° to two decimal places.
5True or false: to use the sine law you need a complete opposite pair.
More practice: on your own
1Solve △ABC (find all unknown parts): ∠A = 38°, ∠C = 80°, c = 25 cm.
∠B = 62°. a = 25 sin 38° ÷ sin 80° ≈ 15.6 cm. b = 25 sin 62° ÷ sin 80° ≈ 22.4 cm.
2In a GeoGebra investigation, Omar's triangle shows a/sin A = 7.43, b/sin B = 7.43 and c/sin C = 7.43. He drags a vertex and the numbers all become 9.12. What does this show?
The ratio depends on the triangle (its size), but within each triangle all three ratios stay equal. This supports the sine law for every triangle he tried.
3Explain why the sine law cannot start if you know only b, c and ∠A.
∠A's opposite side a is unknown, and ∠B and ∠C are unknown, so there is no complete opposite pair. (The cosine law handles this case.)
4A triangular sail has angles 58° and 64° at the two bottom corners. The bottom edge is 3.2 m. Find the two other edges.
Top angle = 58°. Edge opposite 64° = 3.2 sin 64° ÷ sin 58° ≈ 3.39 m. Edge opposite 58° = 3.2 sin 58° ÷ sin 58° = 3.2 m.
5Show that the altitude method also gives b/sin B = c/sin C. (Draw the altitude k from A to BC.)
In the right triangle containing ∠B, k = c sin B. In the right triangle containing ∠C, k = b sin C. So c sin B = b sin C. Divide by sin B sin C: c/sin C = b/sin B.
Challenge
1In acute △ABC, a = 10 and a/sin A = 12. Find ∠A.
sin A = 10/12, so ∠A = sin⁻¹(0.8333) ≈ 56.4°.
2Use the sine law to explain why, in an acute triangle, the longest side is across from the largest angle.
Each side equals the same constant k times the sine of its opposite angle (a = k sin A, etc.). For angles between 0° and 90°, a bigger angle has a bigger sine, so the biggest angle gives the biggest side.
3Use the sine law in a right triangle (∠C = 90°) to show that a = c sin A, which is the SOH ratio.
a/sin A = c/sin 90° = c/1 = c. So a = c sin A, i.e., sin A = a/c = opposite/hypotenuse. The sine law agrees with SOH CAH TOA in right triangles.
Quiz
2. In the derivation of the sine law, the altitude h from C is written as:
3. In a spreadsheet, which formula correctly gives side ÷ sine of angle when the angle in C2 is in degrees?
4. In △ABC, ∠A = 48°, ∠B = 67°, a = 21 m. What is b?
5. You know ∠P, ∠Q and side PQ (between them). What is your FIRST step?
6. Measured ratios in one triangle are 14.0, 14.0 and 13.98. What is the best conclusion?
Big idea: Expectation C1.3 asks students to verify the sine law and the cosine law through investigation (for example, with dynamic geometry software or by measuring drawn triangles; a spreadsheet also works). This lesson does that for the sine law and starts C1.4 (using it to calculate sides).
How to help: If you have a computer, GeoGebra (free at geogebra.org) makes the investigation quick and convincing. Otherwise the paper version works well. Ask your teen to circle the complete opposite pair before writing any equation.
Common difficulties:
- Pairing a side with an angle that is not across from it (Example 3).
- Forgetting to find the third angle first when the known side is between the known angles.
- Spreadsheet SIN without RADIANS gives wrong values.