# Elementary Functions Triangulation and the Law of Sines In ......

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Elementary FunctionsPart 5, Trigonometry

Lecture 5.3a, The Laws of Sines and Triangulation

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 26

Triangulation and the Law of Sines

A triangle has three sides and three angles; their values provide six piecesof information about the triangle.In ordinary geometry, most of the time three pieces of information aresufficient to give us the other three pieces of information.

In order to more easily discuss the angles and sides of a triangle, we willlabel the angles by capital letters (such as A,B,C) and label the sides bysmall letters (such as a, b, c.)

We will assume that a side labeled with a small letter is the side oppositethe angle with the same, but capitalized, letter.For example, the side a is opposite the angle A.

We will also use these letters for the values or magnitudes of these sidesand angles, writing a = 3 feet or A = 14.

Smith (SHSU) Elementary Functions 2013 2 / 26

Three Bits of a Triangle

Suppose we have three bits of information about a triangle. Can werecover all the information?

AAAIf the 3 bits of information are the magnitudes of the 3 angles, and if wehave no information about lengths of sides, then the answer is No.Given a triangle with angles A,B,C (in ordinary Euclidean geometry) wecan always expand (or contract) the triangle to a similar triangle whichhas the same angles but whose sides have all been stretched (or shrunk)by some constant factor.

This Angle-Angle-Angle information (AAA) is not enough informationto describe the entire triangle.

However, we will discover that in almost every other situation, we canrecover the entire triangle.

Smith (SHSU) Elementary Functions 2013 3 / 26

AA and AAA

If we know the values of two angles then since the angles sum to (= 180), we really know all 3 angles.

So AAA is really no better than AA!

We need to know at least one of the sides.

Smith (SHSU) Elementary Functions 2013 4 / 26

ASA and AAS

Suppose we know two angles and one side. If the known side is betweenthe two angles we abbreviate this ASA.

If we know two angles and one side, but the known side is not between thetwo angles, then we are in the Angle-Angle-Side (AAS) situation and wecan also recover the remaining bits of the triangle.

We do this using the Law of Sines.

Smith (SHSU) Elementary Functions 2013 5 / 26

The Law of Sines and ASA

The law of sines says that give a triangle 4ABC, with lengths of sidesa, b and c then

sinA

a=

sinB

b=

sinC

c(1)

Why is this true?

Smith (SHSU) Elementary Functions 2013 6 / 26

The Law of Sines

Start with a triangle with vertices A,B,CDraw a perpendicular from the vertex C onto the side AB and let Drepresent the right angle formed there.If h is the height of the triangle then

sinA =h

band sinB =

h

a.

Smith (SHSU) Elementary Functions 2013 7 / 26

The Law of Sines

sinA = hb and sinB =ha .

Solve for h: h = b sinA and h = a sinB.Since both b sinA and a sinB are equal to h then

b sinA = a sinB.

Dividing both sides by ab we havesinAa =

sinBb .

A similar argument tells us thatsinAa =

sinCc

Smith (SHSU) Elementary Functions 2013 8 / 26

The Law of Sines

sinA

a=

sinB

b=

sinC

c(2)

Smith (SHSU) Elementary Functions 2013 9 / 26

Two Angles and the Law of Sines

If we know two angles of a triangle, then since the three angles add to180 = then we can figure out the third angle. As long as we know onemore bit of information, the length of a side, then the law of sines gives usthe length of all sides.In our abbreviated notation for the known information, this includes thecases ASA and AAS, that is, the cases in which we know two angles andan included side or two angles and a side not between them. These casesare equivalent since, essentially, we know all three angles once we knowtwo.

Smith (SHSU) Elementary Functions 2013 10 / 26

Some Worked Problems.

Use the Law of Sines to solve the following triangles.A = 62, B = 74, c = 14 feet

Solution. If A = 62, B = 74, c = 14 feet then C = 54. Then, by thelaw of sines,

sin 54

14=

sin 62

a=

sin 75

bso

a =14 sin 62

sin 54 15.28 feet and b = 14 sin 75

sin 54 16.72 feet.

Smith (SHSU) Elementary Functions 2013 11 / 26

Some Worked Problems.

Solve the triangle A = 60, B = 70, c = 10 feet

Solution. If A = 60, B = 70, c = 10 feet then C = 50since the sum ofthe angles of a triangle is 180.By the law of sines,

sin 60

a=

sin 70

b=

sin 50

10.

So a =10 sin 60

sin 50= 11.31 feet and b =

10 sin 70

sin 50= 12.27 feet

Smith (SHSU) Elementary Functions 2013 12 / 26

Law of Sines

In the next presentation, we will continue to look at the Law of Sines,investigating the SSA case.

(End)

Smith (SHSU) Elementary Functions 2013 13 / 26

Elementary FunctionsPart 5, Trigonometry

Lecture 5.3b, The Laws of Sines and Triangulation

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 14 / 26

The law of sines and SSA

If we know only one angle of a triangle but two sides, sometimes the Lawof Sines is sufficient.

To use the laws of sines, we need one of the sides to be opposite theknown angle.

In this case, our information is often abbreviated SSA we know twosides and then an angle not between them.

We do have to be a bit careful here. Just because we know the sine of anangle does not mean we know the value of the angle!

There could be two angles with the same sine!

Smith (SHSU) Elementary Functions 2013 15 / 26

Some Worked Problems

Solve the following triangle. (Find all sides and all angles.)A = 30, a = 6 feet, b = 8 feet

Solution. To solve A = 30, a = 6 feet, b = 8 feet apply the law of sines:

sin 30

6 =sinB8 =

sinCc

This forces

sinB = 8 sin 30

6 =23

so B is either sin1(23) 41.81 or B = 180 41.81 = 138.19.

If B = 41.81 then C = 108.19. If B = 138.19 then C = 11.81.By the law of sines c = 6 sinCsin 30 = 12 sinC.

If C = 108.19 then c = 12 sin 108.19 = 11.40But if C = 11.81 then c = 12 sin 11.81 = 2.46

Smith (SHSU) Elementary Functions 2013 16 / 26

Some Worked Problems

We solved the triangle A = 30, a = 6 feet, b = 8 feet:

There are two solutions, two triangles:

A B C a b c

30 41.81 108.19 6 8 11.41

30 138.19 11.81 6 8 2.46

Smith (SHSU) Elementary Functions 2013 17 / 26

Some Worked Problems

Solve the triangle A = 60, a = 10 feet, c = 12 feet

Solutions. If A = 60, a = 10 feet, c = 12 feet then by the law of sines

sin 60

10=

sinC

12

So

sinC =12 sin 60

10=

33

5 1.039 > 1.

There is no angle with sine greater than 1 so there is no solution;

This triangle cannot exist.

Smith (SHSU) Elementary Functions 2013 18 / 26

Some Worked Problems

Solve the triangle a = 10, b = 6, B = 20.

Solution. Use the Law of Sines to see that A could be either A = 34.75

or A = 145.25,

Then C = 125.25 or C = 14.75.

Finally use the Law of Sines to finish off the problem.

There are two answers. Both must be listed.

a = 10, b = 6, c = 14.33, A = 34.75, B = 20, C = 125.25

and

a = 10, b = 6, c = 4.47, A = 145.25, B = 20, C = 14.75.

Smith (SHSU) Elementary Functions 2013 19 / 26

Some Worked Problems

Solve the triangle b = 12, c = 10, C = 60.

Solution. By the Law of Sines, sinB =12

10(sin 60) =

3

5

3 1.039.

Since the sine of B cannot be greater than one,no such triangle is possible.

(The length of b is not large enough to reach the side a.)

Smith (SHSU) Elementary Functions 2013 20 / 26

Some Worked Problems

Solve the triangle b = 12, c = 10, C = 45.

Solution. By the Law of Sines either B 58.06 or B 121.95,

Then A 76.95 or A 13.05.

Finally use the Law of Sines to finish off the problem.

There are two answers. Both must be listed.

a = 13.78, b = 12, c = 10, A = 76.95, B = 58.45, C = 45,

and

a = 3.19, b = 12, c = 10, A = 13.05, B = 121.95, C = 45

Smith (SHSU) Elementary Functions 2013 21 / 26

Triangulation

One can often measure distance to an object by working out a baselineand then measuring the angles formed by lines from the distant object tothe ends of the baseline.

This technique is called triangulation.

This is a classic case of ASA and is perfect for the law of sines.

Smith (SHSU) Elementary Functions 2013 22 / 26

Triangulation

Suppose that a ship is out in the harbor. Standing at the east end of thedock, a compass reveals that the ship is at a bearing of 30 east of north.But if one walks 200 yards west of the dock, the ship is now at a bearingof 40 east of north.

How far is the ship from the dock?

Smith (SHSU) Elementary Functions 2013 23 / 26

Triangulation

Draw an east-west baseline representing the 200 yards and then draw linesfrom the ends of the baseline to the ship. We have a triangle in which thevertex on the west side of the baseline has an (interior) angle of90 40 = 50. The vertex on the east side of the baseline has an angleof 90 + 30 = 120. Call the length of the baseline c = 200, and so wehave angles A = 50 and B = 120.

Smith (SHSU) Elementary Functions 2013 24 / 26

Triangulation

Let us u

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