Law of SinesLaw of Sines

7.17.1

JMerrill, 2009JMerrill, 2009

A Generic TriangleA Generic Triangle

a

b

c

A

B

C

Solve the right Solve the right triangle.triangle.

sin 378

8sin 37

4.815

a

a

a

cos378

8cos37

6.389

b

b

b

90 37 53 B

Solve the right Solve the right triangle.triangle.

1

12sin

1312

sin13

67.380

A

A

A

1

12cos

1312

cos13

22.620

B

B

B

5a

Oblique TrianglesOblique Triangles

But what if the triangle isn’t right? We need a method that will work for what we call oblique triangles.

(Triangles that aren’t right)

Law of SinesLaw of Sines

AA

Usually used when Usually used when you have ASA or you have ASA or AAS AAS

In In ANYANY triangle ABC: triangle ABC:

(Notice that each (Notice that each angle goes with its angle goes with its corresponding side corresponding side in each proportion. in each proportion. One of the One of the proportions must proportions must have a “known” have a “known” angle and side)angle and side)

B C

c b

a

sin A sin B sin C

a b c

Example: Find the Example: Find the missing variablemissing variable In the triangle below, mIn the triangle below, mA = 33A = 33oo, ,

mmB = 47B = 47oo & b = 14, find a. & b = 14, find a.

A B

Csin 47 sin 33

14 asin 47 14sin 33

10.426

a

a

4733

14

You know both B’s, the angle and the side, so that will be our “known”

ExampleExample

A civil engineer A civil engineer wants to determine wants to determine the distances from the distances from points A and B to an points A and B to an inaccessible point C, inaccessible point C, as shown. From as shown. From direct measurement, direct measurement, the engineer knows the engineer knows that AB = 25m, that AB = 25m, A = 110 A = 110oo, and , and B = 20B = 20oo. Find . Find AC and BC. AC and BC.

A

B

C

ExampleExample

1.1. Mark the Mark the drawing with drawing with known info known info

2.2. C = 180 – (110 + 20)C = 180 – (110 + 20)

= 50= 50oo

33..

4.4.

A

B

C

25110110oo

2020oo

sin 110 sin 50

a 25

25sin110 sin 50

30.7

a

a

sin 20 sin 50

b 25

25sin 20 sin 50

11.2

b

b

You DoYou Do

Find ALL missing angles and sides in the Find ALL missing angles and sides in the triangle below: mtriangle below: mA = 28A = 28oo, a = 12 & b , a = 12 & b =24=24

A

B

CmmB = 70B = 70oo

mmC = 82C = 82oo

C = 25.312C = 25.312

Ambiguous CaseAmbiguous Case

When two sides and a non-included angle are given, there are several situations possible – this is called the ambiguous case.

There could be only one triangle.

There could be two triangles (when an angle(s) has more than one possibility).

There could be no triangle (when you take the inverse sine of a value larger than 1).

ExampleExample

Find angle B in triangle ABC if a = Find angle B in triangle ABC if a = 2, b = 6, and A = 302, b = 6, and A = 30oo

Applying the Law of Sines, we Applying the Law of Sines, we have :have :osin30 sinB

2 66sin30 2sinB

6sin30sinB

2 1.5 sinB

Since sinB can never be larger than 1, this triangle does not exist.

Example 2Example 2

Find the missing Find the missing parts in triangle ABC parts in triangle ABC if a = 54cm, b = if a = 54cm, b = 62cm, and A = 4062cm, and A = 40oo

First solve for BFirst solve for B Since B can be 48Since B can be 48oo

or 132or 132oo, C can be , C can be 9292oo or 8 or 8oo

Then c can be 84cm Then c can be 84cm or 12cm (using Law or 12cm (using Law of Sines)of Sines)

o

1 o o

sin40 sinB54 62

62sin40 54sinB

62sin40sinB

54 .7380 sinB

sin B 48 or 132

Thus, there are 2 possibilities and no way to tell which is correct—ambiguous!

Example 3-ApplicationExample 3-Application

A forest ranger at an observation point (A) sights a fire in the direction 32° east of north. Another ranger at a second observation point (B), 10 miles due east of A, sights the same fire 48° west of north. Find the distance from each observation point to the fire.

10

32o

48o

58o 42o

A B

80o 8.6116.795