5.3 The Ambiguous Case If we are given 2 sides and one angle opposite of one of the sides, we may.
Download ReportTranscript 5.3 The Ambiguous Case If we are given 2 sides and one angle opposite of one of the sides, we may.
Slide 1
5.3 The Ambiguous Case
Slide 2
If we are given 2 sides and one angle opposite of one of the sides, we
may not get just one answer.
There are 3 possibilities … one answer, 2 answers or even no answer!
The arrangement of the info is SSA
& we remember this was not
allowed in geometry
Let’s think about this pictorially before trying out the math.
For all these examples, we will consider being given m∠A, a, and b
C
a
b
A
c
B
Slide 3
C
b
A
a
B
c
∠A could be acute… and a < b
b
a
a or
b
a
A
A
2 solutions
if bsinA < a < b
1 solution
if a = bsinA
∠A could be acute… and a ≥ b
b
A
a
a
only 1 solution
b
A
No solution
if a < bsinA
Slide 4
C
b
A
a
B
c
∠A could be obtuse
a
a
b
b
A
A
1 solution
if a > b
No solution
if a ≤ b
Slide 5
Ex 1) Determine the number of solutions.
a) m∠A = 29°, b = 31, a = 23
m∠A is acute
a
so 2 solutions
bsinA = 31sin29° = 15.02 < 23 < 31
a
b
b) m∠A = 132° , b = 96, a = 105
m∠A is obtuse
and a > b
105 > 96
a
b
1 solution
A
Slide 6
Ex 2) Solve △ABC if m∠A = 35.18°, c = 17.8 and a = 11.46
B
sin C
sin 3 5 .1 8
1 7 .8
17.8
A
35.18°
C
11.46
OR
???
sin C
1 1 .4 6
1 7 .8 sin 3 5 .1 8
1 1 .4 6
C
and here?
180 – 63.49 = 116.51
116.51 + 35.18 = 151.69 < 180
Sure!
m∠C = 116.51°
Two Answers!
C sin
1
1 7 .8 sin 3 5 .1 8
1 1 .4 6
C 6 3 .4 9
Slide 7
Ex 2) Solve △ABC if m∠A = 35.18°, c = 17.8 and a = 11.46
B
11.46
17.8
A
35.18°
C
m∠C = 63.49°
m∠B = 81.33°
b
sin 81.33
OR
m∠C = 116.51°
m∠B = 28.31°
11.46
b
sin 35.18
sin 28.31
b = 19.66
11.46
sin 35.18
b = 9.43
Slide 8
Ex 3) Solve △ABC if m∠A = 71.4°, a = 45.3 and b = 51.4
C
51.4
sin B
45.3
5 1 .4
A 71.4°
sin B
B
sin 7 1 .4
4 5 .3
5 1 .4 sin 7 1 .4
4 5 .3
sin B 1 .0 7 5
WAIT!!
sin has to be between –1 & 1, so…
No Triangle exists
Slide 9
Homework
#503 Pg 261 #1, 5, 7, 15, 25, 29, 34, 35, 37
5.3 The Ambiguous Case
Slide 2
If we are given 2 sides and one angle opposite of one of the sides, we
may not get just one answer.
There are 3 possibilities … one answer, 2 answers or even no answer!
The arrangement of the info is SSA
& we remember this was not
allowed in geometry
Let’s think about this pictorially before trying out the math.
For all these examples, we will consider being given m∠A, a, and b
C
a
b
A
c
B
Slide 3
C
b
A
a
B
c
∠A could be acute… and a < b
b
a
a or
b
a
A
A
2 solutions
if bsinA < a < b
1 solution
if a = bsinA
∠A could be acute… and a ≥ b
b
A
a
a
only 1 solution
b
A
No solution
if a < bsinA
Slide 4
C
b
A
a
B
c
∠A could be obtuse
a
a
b
b
A
A
1 solution
if a > b
No solution
if a ≤ b
Slide 5
Ex 1) Determine the number of solutions.
a) m∠A = 29°, b = 31, a = 23
m∠A is acute
a
so 2 solutions
bsinA = 31sin29° = 15.02 < 23 < 31
a
b
b) m∠A = 132° , b = 96, a = 105
m∠A is obtuse
and a > b
105 > 96
a
b
1 solution
A
Slide 6
Ex 2) Solve △ABC if m∠A = 35.18°, c = 17.8 and a = 11.46
B
sin C
sin 3 5 .1 8
1 7 .8
17.8
A
35.18°
C
11.46
OR
???
sin C
1 1 .4 6
1 7 .8 sin 3 5 .1 8
1 1 .4 6
C
and here?
180 – 63.49 = 116.51
116.51 + 35.18 = 151.69 < 180
Sure!
m∠C = 116.51°
Two Answers!
C sin
1
1 7 .8 sin 3 5 .1 8
1 1 .4 6
C 6 3 .4 9
Slide 7
Ex 2) Solve △ABC if m∠A = 35.18°, c = 17.8 and a = 11.46
B
11.46
17.8
A
35.18°
C
m∠C = 63.49°
m∠B = 81.33°
b
sin 81.33
OR
m∠C = 116.51°
m∠B = 28.31°
11.46
b
sin 35.18
sin 28.31
b = 19.66
11.46
sin 35.18
b = 9.43
Slide 8
Ex 3) Solve △ABC if m∠A = 71.4°, a = 45.3 and b = 51.4
C
51.4
sin B
45.3
5 1 .4
A 71.4°
sin B
B
sin 7 1 .4
4 5 .3
5 1 .4 sin 7 1 .4
4 5 .3
sin B 1 .0 7 5
WAIT!!
sin has to be between –1 & 1, so…
No Triangle exists
Slide 9
Homework
#503 Pg 261 #1, 5, 7, 15, 25, 29, 34, 35, 37