Law of Sines

To find the angles of a triangle, we use the Law of sines. So it is useful for solving triangles of any types: Obtuse, Acute or Right-angled Triangle.

  • It can be used to compute the other sides of a triangle when two angles and one side is given.

Or

  • When two sides and one non included angle is given.

So, we use the Sine rule to find unknown lengths or angles of the triangle. It is also called as Sine Rule, Sine Law or Sine Formula. Check out the law of sines formula to know more about it.

Law of Sine Formula:

  • \(\frac{a}{Sin A}=\frac{B}{Sin B}=\frac{c}{Sin C}\)
  • a: b: c = Sin A : Sin B : Sin C
  • \(\frac{a}{b}=\frac{Sin A}{Sin B}\)
  • Similarly, \(\frac{b}{c}=\frac{Sin B}{Sin C}\)

It denotes that if we divide side a by the Sine of \(\angle A\), it is equal to the division of side b by the Sine of \(\angle B\) and also equal to the division of side c by Sine of \(\angle C\).

Or

The sides of a triangle are to one another in the same ratio as the sines of their opposite angles.

Here, Sin A is a number and a is the length.

Law of Sine Proof:

We need a right-angled triangle to prove the above as the trigonometric functions are mostly defined in terms of this type of triangle only.

Law of Sine

Given: A \(\bigtriangleup\)ABC.

Construction: Draw a perpendicular, CD \(\perp\) AB. Then CD = h is the height of the triangle. h, separates the \(\bigtriangleup\) ABC in two right-angled triangles, \(\bigtriangleup\)CDA and \(\bigtriangleup\)CDB.

To Show: \(\frac{a}{b}=\frac{Sin A}{Sin B}\)

Proof: In the \(\bigtriangleup\) CDA,

Sin A= \(\frac{h}{b}\)

And in \(\bigtriangleup\) CDB,

Sin B = \(\frac{h}{a}\)

Therefore, \(\frac{Sin A}{Sin B}=\frac{\frac{h}{a}}{\frac{h}{b}}=\frac{h}{b}\times \frac{a}{h}=\frac{a}{b}\)

And we proved it.

Similarly, we can prove, \(\frac{Sin B}{Sin C}=\frac{b}{c}\) and so on for any pair of angles and their opposite sides.

Practice Example for Law of Sine Formula:

Question: Solve \(\bigtriangleup\) PQR in which \(\angle\)P = 63.5\(^{\circ}\) and \(\angle\)Q = 51.2\(^{\circ}\) and r = 6.3 cm.

Solution: Let’s Calculate the third angle :

  • \(\angle\)B = 180\(^{\circ}\) – 63.5 \(^{\circ}\)- 51.2 \(^{\circ}\)= 65.3\(^{\circ}\)

Now, Let’s calculate the sides:

\(\frac{6.3}{Sin 65.3}=\frac{p}{Sin 63.5}\) (BC= p)

\(p=\frac{6.3\times Sin 63.5}{Sin 65.3}\)

p= 6.21 cm approx(From the table below)

Similarly, \(\frac{6.3}{Sin 65.3}=\frac{q}{Sin 51.2}\)(q= AB)

\(q=\frac{6.3\times Sin51.2}{Sin 65.3}\)

q= 5.40 cm (From the table below)

\(\angle\) C= 65.3\(^{\circ}\)

p=6.21 cm

q=5.40 cm

Get started with Practicing more of Trigonometric problems with BYJU’S.

Trigonometric Table

θ

sin θ

cos θ

tan θ

cot θ

sec θ

csc θ

 

cos θ

sin θ

cot θ

tan θ

csc θ

sec θ

0

1

0

……..

1

……..

90°

0.017

1

0.017

57.29

1

57.299

89°

0.035

0.999

0.035

28.64

1.001

28.654

88°

0.052

0.999

0.052

19.08

1.001

19.107

87°

0.07

0.998

0.07

14.3

1.002

14.336

86°

0.087

0.996

0.087

11.43

1.004

11.474

85°

0.105

0.995

0.105

9.514

1.006

9.567

84°

0.122

0.993

0.123

8.144

1.008

8.206

83°

0.139

0.99

0.141

7.115

1.01

7.185

82°

0.156

0.988

0.158

6.314

1.012

6.392

81°

 

sin θ

cos θ

tan θ

cot θ

sec θ

csc θ

 

cos θ

sin θ

cot θ

tan θ

csc θ

sec θ

10°

0.174

0.985

0.176

5.671

1.015

5.759

80°

11°

0.191

0.982

0.194

5.145

1.019

5.241

79°

12°

0.208

0.978

0.213

4.705

1.022

4.81

78°

13°

0.225

0.974

0.231

4.331

1.026

4.445

77°

14°

0.242

0.97

0.249

4.011

1.031

4.134

76°

15°

0.259

0.966

0.268

3.732

1.035

3.864

75°

16°

0.276

0.961

0.287

3.487

1.04

3.628

74°

17°

0.292

0.956

0.306

3.271

1.046

3.42

73°

18°

0.309

0.951

0.325

3.078

1.051

3.236

72°

19°

0.326

0.946

0.344

2.904

1.058

3.072

71°

 

sin θ

cos θ

tan θ

cot θ

sec θ

csc θ

 

cos θ

sin θ

cot θ

tan θ

csc θ

sec θ

20°

0.342

0.94

0.364

2.747

1.064

2.924

70°

21°

0.358

0.934

0.384

2.605

1.071

2.79

69°

22°

0.375

0.927

0.404

2.475

1.079

2.669

68°

23°

0.391

0.921

0.424

2.356

1.086

2.559

67°

24°

0.407

0.914

0.445

2.246

1.095

2.459

66°

25°

0.423

0.906

0.466

2.145

1.103

2.366

65°

26°

0.438

0.899

0.488

2.05

1.113

2.281

64°

27°

0.454

0.891

0.51

1.963

1.122

2.203

63°

28°

0.469

0.883

0.532

1.881

1.133

2.13

62°

29°

0.485

0.875

0.554

1.804

1.143

2.063

61°

 

sin θ

cos θ

tan θ

cot θ

sec θ

csc θ

 

cos θ

sin θ

cot θ

tan θ

csc θ

sec θ

30°

0.5

0.866

0.577

1.732

1.155

2

60°

31°

0.515

0.857

0.601

1.664

1.167

1.972

59°

32°

0.53

0.848

0.625

1.6

1.179

1.887

58°

33°

0.545

0.839

0.649

1.54

1.192

1.836

57°

34°

0.559

0.829

0.675

1.483

1.206

1.788

56°

35°

0.574

0.819

0.7

1.428

1.221

1.743

55°

36°

0.588

0.809

0.727

1.376

1.236

1.701

54°

37°

0.602

0.799

0.754

1.327

1.252

1.662

53°

38°

0.616

0.788

0.781

1.28

1.269

1.624

52°

39°

0.629

0.777

0.81

1.235

1.287

1.589

51°

 

sin θ

cos θ

tan θ

cot θ

sec θ

csc θ

 

cos θ

sin θ

cot θ

tan θ

csc θ

sec θ

40°

0.643

0.766

0.839

1.192

1.305

1.556

50°

41°

0.656

0.755

0.869

1.15

1.325

1.524

49°

42°

0.669

0.743

0.9

1.111

1.346

1.494

48°

43°

0.682

0.731

0.933

1.072

1.367

1.466

47°

44°

0.695

0.719

0.966

1.036

1.39

1.44

46°

45°

0.707

0.707

1

1

1.414

1.414

45


Practise This Question

A drain cover is made from a square metal plate of side 40 cm and has  434  holes of radius 1 cm each drilled in it. Find the area in cm2 of the remaining square plate.