work out problems on board reminder about minutes/seconds at the end

Work out problems on board Reminder about minutes/seconds at the end

Chapter 6.1

The Law of Sines

We know that Trigonometry can help us solve right triangles. But not all triangles are right triangles. Fortunately, Trigonometry can help us solve non-right triangles as well.

Non-right triangles are know as oblique triangles. There are two categories of oblique triangles—acute and obtuse.

4

Working with Non-right Triangles SOHCAHTOA only works in RIGHT triangles! How can we solve for unknowns in oblique

triangles?

BA

C

a

c

b

5

Working with Non-right Triangles We wish to solve triangles which are not

right triangles

BA

C

a

c

b hsin ? sin ?A B

sin sin

sin sin

sin sin

sin sin

h hA B

b ab A h a B

b A a B

A B

a b

sinC

c Law of Sines!

The Law of Sines is used when we know any two angles and one side (ASA or AAS) or when we know two sides and an angle opposite one of those sides (ASS).

Hints: *Create a proportion with only 1 unknown

Ex:

* Use 4 decimalsknown

x

known

known

Ex 1: ASA.

From the model, we need to determine a, b, and (gamma) using the law of sines.

First off, 42º + 61º + = 180º so that = 77º. (Knowledge of two angles yields the third!)

Now by the law of sines, we have the following relationships:

)sin(42 sin(77 ) sin(61 ) sin(77 ) ;

a 12 b 12

Let’s solve for our unknowns:

o o

o o12sin(42 ) 12sin(61 )

a bsin(77 ) sin(77 )

12(0.6691) 12(0.8746)a b

0.9744 0.9744

a 8.2401 b 10.7709

Ex. 2: AAS

From the model, we need to determine a, b, and using the law of sines.

Note: + 110º + 40º = 180º so that = 30º

ab

By the law of sines, we have the following relationships:

)sin(30 sin(40 ) sin(110 ) sin(40 ) ;

a 12 b 12

Therefore,

12sin(30 ) 12sin(110 )a b

sin(40 ) sin(40 )

12(0.5) 12(0.9397)a b

0.6428 0.6427

a 9.3341 b 17.5428

o o

o o

The Ambiguous Case – ASS

In this case, you may have information that results in one triangle, two triangles, or no triangles.

If I know one of the possible angles, how do I find the other possible angle?

For example, if what solutions of x could be an angle in a triangle?

What do you notice

about these two solutions?

For the case of 2 possible solutions:

2

3sin x

12060

3

2

3

andx

or

andx

Supplementary!

Between 0 and 180 deg.

Or if what solutions of x could be an angle in a triangle?

What do you notice

about these two solutions?

For the case of 2 possible solutions:

2

1sin x

15030

6

5

6

andx

or

andx

Supplementary!

The same is true for all other angle measures with the equivalent sine

So we can subtract the first angle from or 180 degrees to get the second angle.

For the case of 2 possible solutions:

Example #3: ASS

Two sides and an angle opposite one of the sides are given. Let’s try to solve this triangle.

By the law of sines,

sin(57 ) sin( )

15 20

Thus,

20sin(57 )sin( )

1520(0.8387)

sin( )15

sin( ) 1.1183 Impossible!

o

Therefore, there is no value for that exists! No triangle is possible!

Example #4: ASS

Two sides and an angle opposite one of the sides are given. Let’s try to solve this triangle.

By the law of sines,

sin(32 ) sin( )

30 42

So that,

42sin(32 )sin( )

3042(0.5299)

sin( )30

sin( ) 0.7419

48 or 132

o

o o

Interesting! Let’s see if one or both of these angle measures makes sense.

Case 1 Case 2

48 32 180

100

132 32 180

16

Both triangles are valid! Therefore, we have two possible cases to solve.

Finish Case 1: )sin(100 sin(32 )

c 30

30sin(100 )c

sin(32 )

30(0.9848)c

0.5299

c 55.7539

o

o

Finish Case 2: sin(16 ) sin(32 )

c 30

30sin(16 )c

sin(32 )

30(0.2756)c

0.5299

c 15.6029

o

o

Wrapping it up, here are our two solutions:

Example #5: ASS:

Two sides and an angle opposite one of the sides are given. Let’s try to solve this triangle.

By the law of sines,

sin(40 ) sin( )

3 2

2sin(40 )sin( )

32(0.6428)

sin( )3

sin( ) 0.4285

25.4 or 154.6

o

o o

Note: Only one is legitimate!

40 25.4 180

114.6

40 154.6 180

14.6 Not Possible!

Thus, we have only one triangle.

Now let’s find b.

By the law of sines,

sin(114.6 ) sin(40 )

b 3

o o

3sin(114.6 )b

sin(40 )

3(0.9092)b

0.6428

b 4.2433

o

o

Finally, we have:

The Area of a TriangleUsing Trigonometry

Given two sides and the included angle, can we find the area of the triangle?

Remember

1s

2s

bhA2

1

The Area of a TriangleUsing Trigonometry

We can find the area of a triangle if we are given any two sides of a triangle and the measure of the included angle.

(SAS)

1Area = (side)(side)(sine of included angle)

2

Example 7: Find the area ofgiven a = 32 m, b = 9 m, and

ABC36 .m C

132 9 sin36

2Area m m

284.6Area m

Example 6: Finding the Height of a Telephone Pole

A road slopes 15 above the horizontal, and a vertical telephone pole

stands beside the road. The angle of elevation of the Sun is 65 , and

the pole casts a 15 foot shadow downhill along the road. Fin

d the height

of the pole.

x

15ft15º

65º

B

A

C

Let the height of the pole.

180 90 65 25

65 15 50

sin 25 sin50

1515sin50

27.2sin 25

The height of the pole is about 27.2 feet.

o o o o

o o o

x

BAC

ACB

x

x

SKIP???

One min (single prime) is 1/60 of a degree

One second (double prime is 1/60 of a minute or 1/3600 of a degree.

Convert to degrees 40 + (20 * 1/60) + (50 * 1/60 * 1/60)

Angles in Mins/secs

6.1 Pg. 436 #1-7 odd, 13, 14, 19-23 odd, 29, 31, 35, 36

H Dub

6.1 Pg. 436 #1-27 odd, 35

H Dub