Solving triangles: the law of sines and the law of cosines

In this lesson you will learn

Use the law of sines a/sin A = b/sin B = c/sin C = 2R to find sides, angles and the circumradius, and switch between sides and sines.
Use the law of cosines to find a side from two sides and the included angle, or an angle from three sides.
Find the area S = ½ab sin C, decide how many triangles given data describe and whether a triangle is acute, right or obtuse.
Solve measurement problems with angles of elevation and depression, bearings and inaccessible distances.

How wide is a river you cannot cross, or how tall is a tower you cannot climb? Measure one distance and two angles on your side, and trigonometry does the rest. Chinese textbooks call this 解三角形 (solving triangles), and it is a favourite of CSCA-style papers because it links geometry with the values of sin and cos. Notation: in triangle ABC the sides a, b, c lie opposite the angles A, B, C, and R is the radius of the circumscribed circle. The values of special angles and the addition formulas are in the lesson “Trigonometric functions and identities”; the graphs of these functions are in “Graphs of trigonometric functions”.

The law of sines

Definition

Solving a triangle

Finding the unknown sides and angles of a triangle from three known elements, at least one of which is a side. Three angles alone fix only the shape of a triangle, not its size.

a / sin A = b / sin B = c / sin C = 2Ra / sin A = b / sin B = c / sin C = 2R

where:

a, b, cthe sides opposite the angles A, B, C
A, B, Cthe angles of the triangle
Rthe radius of the circumscribed circle (外接圆半径)

Consequences: a = 2R sin A, sin A = a/(2R) and a : b : c = sin A : sin B : sin C. Use it when two angles and a side are known, or two sides and the angle opposite one of them.

Why 2R? Draw the diameter BD of the circumscribed circle. The inscribed angles at A and D stand on the same arc BC, so ∠D = ∠A (or 180° − A, which has the same sine), and ∠BCD = 90° because BD is a diameter. In the right triangle BCD, a = BD · sin D = 2R sin A. A useful consequence: in every triangle the larger side lies opposite the larger angle, a > b ⇔ A > B ⇔ sin A > sin B.

Worked examples: the law of sines

1) In △ABC, A = 30°, B = 45° and a = 2. Find b, C and the circumradius R.
2) a = 3 and A = 60°. Find R.
3) A : B : C = 1 : 2 : 3. Find a : b : c.

Show solution

1) b = a sin B / sin A = 2 · (√2/2) / (1/2) = 2√2; C = 180° − 30° − 45° = 105°; 2R = a / sin A = 2 / (1/2) = 4, so R = 2.
2) 2R = 3 / (√3/2) = 6/√3 = 2√3, so R = √3.
3) A = 30°, B = 60°, C = 90°, so a : b : c = sin 30° : sin 60° : sin 90° = 1/2 : √3/2 : 1 = 1 : √3 : 2 — not 1 : 2 : 3.

The law of cosines

a² = b² + c² − 2bc cos A cos A = (b² + c² − a²) / (2bc)a² = b² + c² − 2bc cos A cos A = (b² + c² − a²) / (2bc)

where:

athe side opposite A
b, cthe two sides that form the angle A
Athe angle between b and c, the included angle (夹角)

The same holds for the other sides: b² = a² + c² − 2ac cos B, c² = a² + b² − 2ab cos C. For A = 90°, cos A = 0 and the law becomes Pythagoras' theorem. The first form finds a side from two sides and the included angle, the second an angle from three sides.

Worked examples: the law of cosines

1) b = 3, c = 8 and A = 60°. Find a.
2) The sides of a triangle are 5, 7 and 8. Find the angle opposite the side 7.
3) In △ABC, b² + c² − a² = bc. Find A.

Show solution

1) a² = 9 + 64 − 2 · 3 · 8 · (1/2) = 73 − 24 = 49, so a = 7.
2) cos θ = (5² + 8² − 7²) / (2 · 5 · 8) = (25 + 64 − 49)/80 = 40/80 = 1/2, so θ = 60°.
3) cos A = (b² + c² − a²)/(2bc) = bc/(2bc) = 1/2, so A = 60°. Compare the given expression with the numerator of the cosine form: that is the whole trick.

CSCA-style item

In △ABC, a = 7, b = 8 and c = 13. What is the largest angle of the triangle?
A) 90° B) 120° C) 135° D) 150°

Show solution

The largest angle lies opposite the largest side c: cos C = (49 + 64 − 169)/(2 · 7 · 8) = −56/112 = −1/2, so C = 120°: B. A would need c² = a² + b², but 169 > 113. C and D would need cos C = −√2/2 or −√3/2.

Area, the number of solutions and the type of a triangle

S = ½ab sin C = ½bc sin A = ½ca sin B

where:

a, btwo sides
Cthe angle between them
Sthe area of the triangle

Half the product of two sides and the sine of the angle between them. Together with the law of sines it also gives S = abc/(4R).

Worked examples: the area

1) a = 4, b = 5 and C = 30°. Find S.
2) The area of △ABC is 4√3, b = c = 4 and A is acute. Find A and a.
3) Find the area of an equilateral triangle with side a.

Show solution

1) S = ½ · 4 · 5 · (1/2) = 5.
2) sin A = 2S/(bc) = 8√3/16 = √3/2, and A is acute, so A = 60° (an obtuse A would give 120°). Then a² = 16 + 16 − 2 · 4 · 4 · (1/2) = 16, a = 4: the triangle is equilateral.
3) S = ½ · a · a · sin 60° = √3a²/4.

CSCA-style item

In △ABC, B = 60°, a = 4 and the area is 2√3. What is b?
A) 2 B) 2√3 C) 2√7 D) 4

Show solution

S = ½ac sin B = ½ · 4 · c · (√3/2) = √3c = 2√3, so c = 2. Then b² = 16 + 4 − 2 · 4 · 2 · (1/2) = 12 and b = 2√3: B. A is c, not b. C adds 2ac cos B instead of subtracting it (16 + 4 + 8 = 28). D is a; in fact b² + c² = 12 + 4 = 16 = a², so the triangle is right-angled at A.

When two sides and an angle opposite one of them are given (a, b and A), the data may describe no triangle, one or two. Draw the angle A and the side b = AC; the third vertex B lies on the other side of the angle at the distance a from C. The circle of radius a around C can miss that side, touch it or cross it twice. Everything depends on the height h = b sin A from C.

Here h < a < b: two triangles. With a < h the arc does not reach the side, with a = h it touches it, and with a ≥ b the second crossing falls behind A, so one triangle remains.

h = b sin A

where:

hthe distance from C to the other side of the angle A (the height)
athe side opposite A
bthe side next to A

A acute: a < h — no triangle; a = h — one right triangle; h < a < b — two triangles; a ≥ b — one triangle. A right or obtuse: one triangle if a > b, none if a ≤ b.

Worked examples: how many triangles?

Find the number of triangles:
1) A = 30°, b = 4, a = 3;
2) A = 60°, b = 2, a = √3;
3) A = 45°, b = 2, a = 3;
4) A = 30°, b = 3, a = 1.

Show solution

1) h = 4 · 1/2 = 2 < 3 < 4: two. Indeed sin B = 4 · (1/2)/3 = 2/3, and both the acute B and the obtuse 180° − B fit.
2) h = 2 · (√3/2) = √3 = a: one, with B = 90°.
3) a = 3 ≥ b = 2: one; B < A, so B is acute.
4) h = 3/2 > 1 = a: none — the law of sines would give sin B = 3/2 > 1.

c² < a² + b² ⇔ C < 90° c² = a² + b² ⇔ C = 90° c² > a² + b² ⇔ C > 90°

where:

cthe longest side
Cthe largest angle (opposite c)

It follows from cos C = (a² + b² − c²)/(2ab): the sign of the numerator is the sign of cos C. The triangle is acute if its largest angle is acute, so check only the longest side.

Worked examples: the type of a triangle

1) The sides are 4, 5 and 6. Is the triangle acute, right or obtuse?
2) The same for the sides 2, 3 and 4.
3) In △ABC, a = 2b cos C. What kind of triangle is it?

Show solution

1) 6² = 36 < 4² + 5² = 41: acute.
2) 4² = 16 > 2² + 3² = 13: obtuse.
3) Replace cos C by the law of cosines: a = 2b · (a² + b² − c²)/(2ab) = (a² + b² − c²)/a, so a² = a² + b² − c² and b = c: the triangle is isosceles.

Measurement problems

Definition

Angle of elevation, angle of depression, bearing

The angle of elevation (仰角) is the angle between the horizontal and the line of sight up to an object; the angle of depression (俯角) is the angle between the horizontal and the line of sight down to it. A bearing such as N 30° E (北偏东30°) means: face north, then turn 30° towards the east.

1
Draw and label
Sketch the situation from above (bearings) or from the side (heights) and put every known length and angle on the sketch.
2
Find a triangle with three known elements
Angles along a straight line add up to 180°, and so do the angles of a triangle; the third angle is often the key.
3
Law of sines or cosines
Two angles and a side: the law of sines. Two sides and the angle between them: the law of cosines.
4
Finish in a right triangle
A height or a horizontal distance usually comes from sin or cos in a right triangle at the end.

Worked examples: heights and distances

1) A tower stands on level ground. From point A the angle of elevation of its top T is 30°; after walking 20 m straight towards the tower to point B, the angle of elevation is 60°. Find the height of the tower.
2) Points A and B lie on one bank of a river, AB = 60 m. A tree C on the other bank is seen with ∠CAB = 45° and ∠CBA = 105°. Find BC.
3) Two ships leave a port at the same time on courses that form an angle of 120°. After two hours one has sailed 30 km and the other 50 km. How far apart are they?

Show solution

1) In △ABT: ∠A = 30°, ∠ABT = 180° − 60° = 120°, so ∠ATB = 30° and the triangle is isosceles: BT = AB = 20 m. The height is BT · sin 60° = 10√3 m ≈ 17.3 m.
2) ∠C = 180° − 45° − 105° = 30°. BC lies opposite A: BC = AB · sin A / sin C = 60 · (√2/2) / (1/2) = 60√2 m ≈ 84.9 m.
3) d² = 30² + 50² − 2 · 30 · 50 · cos 120° = 900 + 2500 + 1500 = 4900, d = 70 km.

CSCA-style item

A ship at point A sees a lighthouse L on a bearing of N 15° E. The ship sails 20 km due north to point B, and from B the lighthouse is on a bearing of N 45° E. How far was the ship from the lighthouse at A?
A) 10√2 km B) 20 km C) 20√2 km D) 40 km

Show solution

In △ABL: ∠A = 15°, and ∠ABL = 180° − 45° = 135° (the angle between the direction back to A, due south, and the direction to L). So ∠L = 180° − 15° − 135° = 30°. AL lies opposite B: AL = AB · sin 135° / sin 30° = 20 · (√2/2) / (1/2) = 20√2 km: C. B assumes an isosceles triangle, A turns the ratio upside down, and D uses sin 90° instead of sin 135°.

How CSCA asks about this

Triangle questions are common, both as direct calculations and in the form “what kind of triangle is it”. Learn the key terms in English and Chinese:

Term	中文	Pinyin
law of sines	正弦定理	zhèngxián dìnglǐ
law of cosines	余弦定理	yúxián dìnglǐ
solving a triangle	解三角形	jiě sānjiǎoxíng
circumscribed circle	外接圆	wàijiēyuán
included angle	夹角	jiājiǎo
opposite side	对边	duìbiān
area	面积	miànjī
acute / obtuse triangle	锐角 / 钝角三角形	ruìjiǎo / dùnjiǎo sānjiǎoxíng
right triangle	直角三角形	zhíjiǎo sānjiǎoxíng
isosceles / equilateral triangle	等腰 / 等边三角形	děngyāo / děngbiān sānjiǎoxíng
angle of elevation	仰角	yǎngjiǎo
angle of depression	俯角	fǔjiǎo
bearing	方位角	fāngwèijiǎo
number of solutions	解的个数	jiě de gèshù

Typical question patterns and time-savers (about 75 s per item):

Find a side or an angle: two angles and a side → sines; two sides and the included angle → cosines; three sides → the cosine of the largest angle.
Expressions like b² + c² − a² = bc: compare with 2bc cos A at once: cos A = 1/2.
Ratios of sines: sin A : sin B : sin C = a : b : c; a ratio of angles is not a ratio of sides.
Area: S = ½ab sin C; if S and two sides are given, sin C = 2S/(ab) — mind the two angles with the same sine.
How many triangles: compare a with h = b sin A and with b.
Type of triangle: the sign of a² + b² − c² for the longest side c; rewrite equations such as a cos A = b cos B with sines.
Measurement: find the third angle first; heights end in a right triangle.

Key points

Law of sines: a/sin A = b/sin B = c/sin C = 2R; the larger side lies opposite the larger angle.
Law of cosines: a² = b² + c² − 2bc cos A; cos A = (b² + c² − a²)/(2bc) finds an angle uniquely.
Area: S = ½ab sin C, with C the angle between the two sides.
Two sides and an opposite angle: compare a with h = b sin A and with b — 0, 1 or 2 triangles.
Type: the sign of a² + b² − c² for the longest side c tells acute, right or obtuse.
Measurement problems: find the third angle, use the law of sines or cosines, finish in a right triangle.

Check yourself

12 questions. Every correct answer earns XP.

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In △ABC, what does a/sin A equal (R is the radius of the circumscribed circle)?

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