LAW of SINES

Definition: Oblique Triangles An oblique triangle is a triangle that has no right angles. C B A a b c To solve an oblique triangle, you need to know the measure of at least one side and the measures of any other two parts of the triangle – two sides, two angles, or one angle and one side. Definition: Oblique Triangles

Solving Oblique Triangles The following cases are considered when solving oblique triangles. Two angles and any side (AAS or ASA) A C c A B c 2. Two sides and an angle opposite one of them (SSA) C c a 3. Three sides (SSS) a c b c a B 4. Two sides and their included angle (SAS) Solving Oblique Triangles

Definition: Law of Sines The first two cases can be solved using the Law of Sines. (The last two cases can be solved using the Law of Cosines.) Law of Sines If ABC is an oblique triangle with sides a, b, and c, then C B A b h c a C B A b h c a Acute Triangle Obtuse Triangle Definition: Law of Sines

Example: Law of Sines - ASA Example (ASA): Find the remaining angle and sides of the triangle. C B A b c 60 10 a = 4.5 ft The third angle in the triangle is A = 180 – C – B = 180 – 10 – 60 = 110. 4.15 ft 110 0.83 ft Use the Law of Sines to find side b and c. Example: Law of Sines - ASA

Example: Single Solution Case - SSA Example (SSA): Use the Law of Sines to solve the triangle. A = 110, a = 125 inches, b = 100 inches C B A b = 100 in c a = 125 in 110 21.26 48.74 48.23 in C  180 – 110 – 48.74 = 21.26 Example: Single Solution Case - SSA

Example: No-Solution Case - SSA Example (SSA): Use the Law of Sines to solve the triangle. A = 76, a = 18 inches, b = 20 inches C A B b = 20 in a = 18 in 76 There is no angle whose sine is 1.078. There is no triangle satisfying the given conditions. Example: No-Solution Case - SSA

Law of Sines: The Ambiguous Case Given: lengths of two sides and the angle opposite one of them (S-S-A) /ctr

Possible Outcomes Case 1: If A is acute and a > b C b a One SOLUTION B A c Solve normally using the Law of Sines

Possible Outcomes Case 2: If A is acute and a < b a. If a < b sinA a C b a h = b sin A b B A h c A B c NO SOLUTION

Possible Outcomes Case 2: If A is acute and a < b b. If a = b sinA h = b sin A b h = a A c B 1 SOLUTION

Possible Outcomes Case 2: If A is acute and a < b h = b sin A b. If a > b sinA C b a h a A B B c 2 SOLUTIONS

Possible Outcomes Case 3: If A is obtuse and a > b C a b A c B ONE SOLUTION

Possible Outcomes Case 3: If A is obtuse and a ≤ b C a b A c B NO SOLUTION

EXAMPLE 1 Given: ABC where a = 22 inches b = 12 inches a>b mA = 42o a>b mA > mB SINGLE–SOLUTION CASE (acute) Find m B, m C, and c.

sin A = sin B a b Sin B  0.36498 mB = 21.41o or 21o Sine values of supplementary angles are equal. The supplement of B is B2.  mB2=159o

mC = 180o – (42o + 21o) mC = 117o sin A = sin C a c c = 29.29 inches SINGLE–SOLUTION CASE

EXAMPLE 2 Given: ABC where c = 15 inches b = 25 inches c < b mC = 85o c < b 15 < 25 sin 85o c ? b sin C NO SOLUTION CASE (acute) Find m B, m C, and c.

sin A = sin B a b Sin B  1.66032 mB = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1  sin   1 NO SOLUTION CASE /ctr

EXAMPLE 3 Given: ABC where b = 15.2 inches a = 20 inches b < a mB = 110o b < a NO SOLUTION CASE (obtuse) Find m B, m C, and c.

sin A = sin B a b Sin B  1.23644 mB = ? Sin B > 1 NOT POSSIBLE ! Recall: – 1  sin   1 NO SOLUTION CASE /ctr

EXAMPLE 4 Given: ABC where a = 24 inches b = 36 inches a < b mA = 25o a < b a ? b sin A 24 > 36 sin 25o TWO – SOLUTION CASE (acute) Find m B, m C, and c.

sin A = sin B a b Sin B  0.63393 mB = 39.34o or 39o The supplement of B is B2.  mB2 = 141o mC1 = 180o – (25o + 39o) mC1 = 116o mC2 = 180o – (25o+141o) mC2 = 14o

sin A = sin C a c1 c1 = 51.04 inches sin A = sin C a c2 /ctr

EXAMPLE 4 Final Answers: mB1 = 39o mC1 = 116o c1 = 51.04 in. TWO – SOLUTION CASE /ctr

Find m B, m C, and c, if they exist. PRACTICE Find m B, m C, and c, if they exist.  1) a = 9.1, b = 12, mA = 35o  2) a = 25, b = 46, mA = 37o 3) a = 15, b = 10, mA = 66o   /ctr

Answers: 1)Case 1: mB=49o,mC=96o,c=15.78 Case 2: 2)No possible solution. 3)mB=38o,mC=76o,c=15.93   /ctr

Area of a Triangle The area A of a triangle is where b is the base and h is an altitude drawn to that base

Area of SAS Triangles If we know two sides a and b and the included angle C, then The area A of a triangle equals one-half the product of two of its sides times the sine of their included angle. Note: bsinC=h

Area of SAS Triangles Example. Find the area A of the triangle for which a = 12, b = 15 and C = 52o