LAW OF SINES: THE AMBIGUOUS CASE. Review Identify if the given oblique triangle can be solved using...
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Transcript of LAW OF SINES: THE AMBIGUOUS CASE. Review Identify if the given oblique triangle can be solved using...
LAW OF SINES:
THE AMBIGUOUS CASE
Review Identify if the given oblique triangle can be
solved using the Law of Sines or the Law of Cosines
1. X = 210, Z = 650 and y = 34.7
2. s = 73.1, r = 93.67 and T = 650
3. a = 78.3, b = 23.5 and c = 36.8
Law of Sines
Law of Cosines
Law of Cosines
Law of Sines: The Ambiguous Case
Given:
lengths of two sides and the angle opposite one of them (S-S-A)
AMBIGUOUS
• Open to various interpretations
• Having double meaning
• Difficult to classify, distinguish, or comprehend
Always set your triangle up this way…
Given angle here
Side opposite hereOther g
iven si
de here
Possible OutcomesCase 1: If is acute and the side opposite the given angle < the other given side.
A
C
B
b a
c
h = b sin A
a. If a < h
A
C
B
b
a
c
h
NO SOLUTION
Possible OutcomesCase 1: If is acute and the side opposite the given angle < the other given side.
A
C
B
b a
c
h = b sin A
b. If a = h
A
C
B
b= a
c
h
1 SOLUTION
Possible OutcomesCase 1: If is acute and the side opposite the given angle < the other given side.
A
C
B
b a
c
h = b sin A
b. If a > h
A
C
B
b
c
h
2 SOLUTIONS
a a
B180 -
Possible OutcomesCase 2: If is obtuse and the side opposite the given angle > the other given side.
C
AB
ab
c
ONE SOLUTION
Possible OutcomesCase 2: If is obtuse and the side opposite the given angle the other given side.
C
A B
a
b
c
NO SOLUTION
SUMMARY is acute or is obtuse
Side opposite < other side
Side opposite > other side
Side opposite > other side
Side opposite < other side
FIND HEIGHT: h = other side sin
Side opposite < h:
Side opposite = h:
Side opposite > h:
No Solution
1 Solution
2 Solutions
1 Solution
1 Solution
No Solution
Given: ABC where
a = 22 inches
b = 12 inches
mA = 42o
EXAMPLE 1
Find m B, m C, and c.(acute)
a>b
mA > mBSINGLE–SOLUTION CASE
sin A = sin B a b
Sin B 0.36498 mB = 21.41o or 21o
Sine values of supplementary angles are equal.
The supplement of B is B2. mB2=159o
mC = 180o – (42o + 21o) mC = 117o
sin A = sin C a c
c = 29.29 inches
SINGLE–SOLUTION CASE
Given: ABC where
c = 15 inches
b = 25 inches
mC = 85o
EXAMPLE 2
Find m B, m C, and c.(acute)
c < b
c ? b sin C 15 < 25 sin 85o
NO SOLUTION CASE
sin A = sin B a bSin B 1.66032 mB = ?
Sin B > 1 NOT POSSIBLE !
Recall: – 1 sin 1
NO SOLUTION CASE
Given: ABC where
b = 15.2 inches
a = 20 inches
mB = 110o
EXAMPLE 3
Find m B, m C, and c.(obtuse)
b < a
NO SOLUTION CASE
sin A = sin B a b
Sin B 1.23644 mB = ?
Sin B > 1 NOT POSSIBLE !
Recall: – 1 sin 1
NO SOLUTION CASE
Given: ABC where
a = 24 inches
b = 36 inches
mA = 25o
EXAMPLE 4
Find m B, m C, and c.(acute)
a < b
a ? b sin A 24 > 36 sin 25o
TWO – SOLUTION CASE
sin A = sin B a b
Sin B 0.63393 mB = 39.34o or 39o
The supplement of B is B2. mB2 = 141o
mC1 = 180o – (25o + 39o) mC1 = 116o mC2 = 180o – (25o+141o) mC2 = 14o
sin A = sin C a c1
c1 = 51.04 inches
sin A = sin C a c2
c = 13.74 inches
Final Answers:
mB1 = 39o
mC1 = 116o
c1 = 51.04 in.
EXAMPLE 3
TWO – SOLUTION CASE
mB2 = 141o
mC2 = 14o
C2= 13.74 in.
SEATWORK: (notebook)
Answer in pairs.
Find m B, m C, and c, if they exist.
1) a = 9.1, b = 12, mA = 35o
2) a = 25, b = 46, mA = 37o
3) a = 15, b = 10, mA = 66o
Answers:
1)Case 1:
mB=49o,mC=96o,c=15.78
Case 2:
mB=131o,mC=14o,c=3.84
2)No possible solution.
3)mB=38o,mC=76o,c=15.93