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The Ambiguous Case of the Law of Sines

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The Ambiguous Case of the Law of Sines. This is the SSA case of an oblique triangle. SSA means you are given two sides and the angle opposite one of them ( a , b , and A ). . First, consider the case where A is an acute angle . b. A. Here are A and side b of fixed lengths .

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Presentation Transcript
slide2

This is the SSA case of an oblique triangle.

SSA means you are given two sides and the angle opposite one of them (a, b, and A).

slide4

b

A

Here are A and side b of fixed lengths.

slide5

b

A

You will be given instructions to make side a of various lengths as we work through the activity.

slide6

We need to look at four possibilities for a, bycomparing a to b and to h (the height of the triangle).

slide7

Look at the case where a >b.

(This would also mean that a >h.)

slide8

b

A

slide9

b

A

One oblique triangle can be made.

C

a

c

B

slide12

1. a < h

2. a = h

3. a > h

slide14

b

h

A

slide15

b

h

A

No triangle is possible.

a

slide16

b

h

A

slide17

b

h

A

One triangle can be made . . .

C

. . . a right triangle.

a

c

B

slide18

b

h

A

slide19

b

h

A

One triangle can be made. . .

C

. . . and angle B is an acute angle .

a

c

B

slide20

b

h

A

Can another triangle be made with this same length?

slide21

b

h

A

Yes, another triangle can be made . . .

C

. . . and angle B is an obtuse angle.

a

c

B

slide23

Look at this right triangle.

Can you write a trig ratio that would give the value for h?

b

h

A

slide24

then

Since

slide27

If a > b

one obliquetriangle

can be formed.

slide28

If a < b

zero, one, or two triangles can be formed.

slide29

Calculate h using

and compare a to it.

slide30

Then choose the appropriate case below.

1. Ifa < h

notriangle can be formed.

2. Ifa = h

one righttriangle can be formed.

3. Ifa >h

two obliquetriangles can be formed.

slide33

b

h

A

Go back to the last two triangles you made on your worksheet.

C

a

c

B

slide34

b

h

A

On a piece of patty paper, trace angle B from this triangle.

C

a

c

B

slide36

b

h

A

Adjacent to angle B from the previous triangle (on the patty paper), trace angle B from this triangle. (You may have to rotate or flip the patty paper.)

C

a

c

B

slide38

What did you find?

The two angles (B) in the two triangles are supplementary.

slide39

So if you could find one of the values for B, you could find the other value by subtracting from 180°.

slide42

A = 36°, a = 16, b = 17

Since a < b, then we have to compare a to h.

slide43

A = 36°, a = 16, b = 17

Since a > h, then there are two triangles.

slide44

A = 36°, a = 16, b = 17

Use the Law of Sines to find B:

slide46

We now know this much about the triangle:

A = 36° a = 16

B ≈39° b = 17

slide47

Now find C.

A + B + C= 180°

C = 180° - (A + B)

C = 180° - (36° + 39°)

C ≈105°

slide48

We now know every part of the triangle except c.

A = 36° a = 16

B ≈39° b = 17

C ≈105° c = ?

slide51

We have now solved one of the triangles.

A = 36° a = 16

B ≈39° b = 17

C ≈105° c ≈ 26

slide52

But remember that there are two triangles.

Also remember that B from the first triangle and B from the second triangle are supplementary.

slide53

Since we found the measure of B in the first triangle to equal 39°, we can find the measure of B in thesecond by subtracting from 180°.

B2 = 180° - B1

B2 = 180° - 39° ≈ 141°

slide54

Here is where we are:

A1 = 36° a1 = 16

B1≈39° b1 = 17

C1≈105° c1≈26

A2 = 36° a2 = 16

B2≈ 141° b2 = 17

C2 = ? c2 = ?

slide55

We can find C2 like we found C1.

A2 + B2 + C2 = 180°

C2 = 180° - (A2 + B2)

C2 = 180° - (36° + 141°)

C2≈3°

slide58

Now we have solved both triangles:

A1 = 36° a1 = 16

B1≈ 39° b1 = 17

C1≈ 105° c1≈26

A2 = 36° a2 = 16

B2≈ 141° b2 = 17

C2≈ 3° c2≈1

slide60

b

A

slide61

b

A

C

slide62

b

A

No triangle is possible.

a

slide63

b

A

slide64

b

A

slide65

b

A

One triangle can be made.

C

a

c

B

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