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 = 21 0, Z = 65 0 and y = s = 73.1, r = and T = a = 78.3, b = 23.5 and c = 36.8 Law of Sines 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 here Other given side here
Possible Outcomes Case 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 Outcomes Case 1: If is acute and the side opposite the given angle < the other given side. AC B b a c h = b sin A b. If a = h A C B b = a c h 1 SOLUTION
Possible Outcomes Case 1: If is acute and the side opposite the given angle < the other given side. AC B b a c h = b sin A b. If a > h A C B b c h 2 SOLUTIONS aa B
Possible Outcomes Case 2: If is obtuse and the side opposite the given angle > the other given side. C A B a b c ONE SOLUTION
Possible Outcomes Case 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 FIND HEIGHT: h = other side sin Side opposite < h: Side opposite = h: Side opposite > h: No Solution 1 Solution 2 Solutions 1 Solution No Solution
Given: ABC where a = 22 inches b = 12 inches m A = 42 o EXAMPLE 1 Find m B, m C, and c. (acute) a>b m A > m B SINGLE–SOLUTION CASE
sin A = sin B a b Sin B m B = o or 21 o Sine values of supplementary angles are equal. The supplement of B is B 2. m B 2 =159 o
m C = 180 o – (42 o + 21 o ) m C = 117 o sin A = sin C a c c = inches SINGLE–SOLUTION CASE
Given: ABC where c = 15 inches b = 25 inches m C = 85 o EXAMPLE 2 Find m B, m C, and c. (acute) c < b c ? b sin C 15 < 25 sin 85 o NO SOLUTION CASE
sin A = sin B a b Sin B m B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1 sin 1 NO SOLUTION CASE
Given: ABC where b = 15.2 inches a = 20 inches m B = 110 o EXAMPLE 3 Find m B, m C, and c. (obtuse) b < a NO SOLUTION CASE
sin A = sin B a b Sin B m B = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1 sin 1 NO SOLUTION CASE
Given: ABC where a = 24 inches b = 36 inches m A = 25 o EXAMPLE 4 Find m B, m C, and c. (acute) a < b a ? b sin A24 > 36 sin 25 o TWO – SOLUTION CASE
sin A = sin B a b Sin B m B = o or 39 o The supplement of B is B 2. m B 2 = 141 o m C 1 = 180 o – (25 o + 39 o ) m C 1 = 116 o m C 2 = 180 o – (25 o +141 o ) m C 2 = 14 o
sin A = sin C a c 1 c 1 = inches sin A = sin C a c 2 c = inches
Final Answers: m B 1 = 39 o m C 1 = 116 o c 1 = in. EXAMPLE 3 TWO – SOLUTION CASE m B 2 = 141 o m C 2 = 14 o C 2 = in.
SEATWORK: (notebook) Answer in pairs. Find m B, m C, and c, if they exist. 1 ) a = 9.1, b = 12, m A = 35 o 2) a = 25, b = 46, m A = 37 o 3) a = 15, b = 10, m A = 66 o
Answers: 1 )Case 1: m B=49 o,m C=96 o,c=15.78 Case 2: m B=131 o,m C=14 o,c=3.84 2)No possible solution. 3)m B=38 o,m C=76 o,c=15.93