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CAMBRIDGE O LEVEL · 2025–2027

Non-right-angled triangles.

Match opposite pairs, use the included angle and check whether the information allows two triangles.

Syllabus 6.3Visual notesWorked examples
BY THE END

Use sine and cosine rules · handle obtuse angles and SSA ambiguity · calculate 12ab sin C areas

01

Pair each side with its opposite angle

The basic sine, cosine and tangent ratios work directly in right triangles. A general triangle may have no right angle, so it needs relationships that link sides and angles without that restriction.

Label each side opposite its matching capital-letter angle. Then inventory the given information. A known opposite side–angle pair points towards the sine rule; two sides with their included angle, or all three sides, point towards the cosine rule. The information chooses the method.

The sine and cosine rules apply to any triangle, including obtuse triangles. Match lowercase side letters with their opposite uppercase angles before substituting.

ABCabc
In triangle ABC, side a is opposite A, side b opposite B, and side c opposite C. Angle C is included between sides a and b.
Choose the rule from the information available
Known informationUseful method
An opposite side–angle pair and another side or angleSine rule; check the ambiguous case when finding an angle
Two sides and the included angle (SAS)Cosine rule for the third side
All three sides (SSS)Cosine rule for an angle
Two sides and their included angleArea = 12ab sin C
Pause and explainWhat does “side a is opposite angle A” mean?

Side a is the side not touching vertex A; it faces angle A across the triangle.

02

Sine rule: keep the opposite pairs together

The sine rule compares a side with the sine of its opposite angle. Its power comes from the fact that this ratio is the same for all three opposite pairs in one triangle.

Find one complete known side–angle pair and compare it with the pair containing your unknown. Keep each side matched with its opposite angle. When finding an angle, remember that an acute and an obtuse angle can share a sine value, so check the triangle’s remaining information.

asin A = bsin B = csin C
WORKED EXAMPLE

In a triangle, A = 40°, B = 65° and a = 8 cm. Find b.

bsin 65° = 8sin 40°
b = 8 sin 65°sin 40° ≈ 11.3 cm

Side a is paired with 40°, and side b with 65°. The third angle is 75°.

Pause and explainWhy is an opposite side–angle pair important when using the sine rule?

It supplies a known ratio that can be equated to the ratio containing the unknown.

03

Cosine rule: sides and the angle between them

Pythagoras describes a triangle when its included angle is 90°. The cosine rule extends that side relationship to other included angles by adding a correction involving cosine.

For a missing side, use the two known sides and the angle between them. For a missing angle, rearrange using all three sides. The side opposite the chosen angle must occupy the matching position in the formula; a careful labelled sketch prevents mismatches.

c² = a² + b² − 2ab cos C
cos C = a² + b² − c²2ab
WORKED EXAMPLE

Two sides are 7 cm and 9 cm with included angle 60°. Find the third side.

c² = 7² + 9² − 2(7)(9) cos 60° = 67
c = √67 ≈ 8.19 cm
WORKED EXAMPLE

A triangle has a = 13 cm, b = 8 cm and c = 7 cm. Find A.

cos A = 8² + 7² − 13²2 × 8 × 7 = −12
A = cos−1(−12) = 120.0°

A negative cosine correctly gives an obtuse angle. The longest side, 13 cm, is opposite the largest angle.

Pause and explainWhat happens to the cosine-rule correction when the included angle is 90°?

Cos 90° is zero, so the correction disappears and the formula becomes Pythagoras.

04

SSA can give two possible triangles

Knowing two sides and a non-included angle does not always fix one triangle. In some arrangements, the remaining vertex can sit in two positions while satisfying the same information.

The calculator’s inverse sine normally gives the acute candidate first. A second candidate is 180° minus that angle. Test each against the given angle and the triangle angle sum; retain only candidates that leave a positive third angle and satisfy the side information.

If the sine rule gives sin B = k, an acute calculator answer B₁ may have a second solution B₂ = 180° − B₁. Test each against A + B < 180°.

WORKED EXAMPLE

Given A = 30°, a = 7 and b = 10, find the possible angles B.

sin B = 10 sin 30°7 = 57
B₁ ≈ 45.6°, B₂ ≈ 134.4°

Both are possible: their remaining angles are approximately 104.4° and 15.6°. Retain full calculator values when finding another side.

AMBIGUOUS CASE · A = 30°, a = 7, b = 10
ACB₁B₂b = 1030°a = 7
Two possible triangles share angle A = 30 degrees and side AC = 10. Both possible vertices B lie on the same ray and are 7 units from C.

B = 45.6°, C = 104.4°.

c = 7 sin Csin 30° ≈ 13.6

The larger triangle has an acute angle B₁.

If the calculated sine is greater than 1, no triangle is possible. If B₁ = 90°, the two candidates coincide. If a candidate makes the angle sum reach or exceed 180°, reject it.

Pause and explainIf one candidate angle is 40°, what is the other angle with the same sine?

140°. It is a possible candidate only if it fits the rest of the triangle.

05

Area: use the included angle

The familiar triangle area is half base times perpendicular height. If the height is not given but an included angle is known, trigonometry can express that height using one of the sides.

This gives the area rule involving two sides and the sine of the angle between them. Make sure the angle is the included angle of the two lengths used. An obtuse angle is allowed; its sine still gives the appropriate positive perpendicular-height contribution.

Area = 12ab sin C
WORKED EXAMPLE

Two sides are 8 cm and 11 cm and their included angle is 125°. Find the area.

Area = 12 × 8 × 11 × sin 125°
≈ 36.0 cm²

The sine of an obtuse triangle angle is positive. Use the included angle; the side opposite that angle is not one of the two multiplied sides.

Pause and explainWhich angle belongs in 12ab sin C?

The angle C between sides a and b, which is opposite side c.

06

Your turn. Choose the rule before substituting.

Make a labelled side–angle inventory before selecting a rule. Keep opposite pairs together and distinguish an included angle from a non-included one.

For inverse-sine answers, check whether a second valid triangle exists. For any calculated angle, verify that the total can still be 180° and that the largest side faces the largest angle.

1

A = 30°, B = 45° and a = 6 cm. Find b.

Reveal worked answer
b = 6 sin 45° ÷ sin 30° = 6√2 cm ≈ 8.49 cm.
2

Two sides are 5 cm and 8 cm with included angle 60°. Find the third side.

Reveal worked answer
c² = 25 + 64 − 80 cos 60° = 49, so c = 7 cm.
3

A triangle has sides 5, 5 and 8 cm. Find the angle opposite the 8 cm side.

Reveal worked answer
cos C = (25 + 25 − 64)(2 × 5 × 5) = −0.28. C ≈ 106.3°.
4

Find the area with sides 10 cm and 12 cm and included angle 30°.

Reveal worked answer
Area = 12 × 10 × 12 × sin 30° = 30 cm².
5

A = 30°, a = 6 and b = 8. Find both possible values of B.

Reveal worked answer
sin B = 8 sin 30° ÷ 6 = 23. B₁ ≈ 41.8° and B₂ ≈ 138.2°. Both leave a positive third angle, so both are valid.