# The Law of Cosines The law of sines cannot be used to solve a triangle where two sides and the included angle (SAS) are given or three sides (SSS) are.

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The Law of Cosines The law of sines cannot be used to solve a triangle where two sides and the included angle (SAS) are given or three sides (SSS) are given . In such cases we use the cosine rule which is defined as follows: If b and c are two sides of a triangle and A, the included angle of these sides, then: C a b a2 = b2 + c2 – 2 bc cos A B c A If a and b are two sides and C the included angle: c2 = a2 + b2 – 2 ab cos C If a and c are two sides and B the included angle: b2 = a2 + c2 – 2 ac cos B

Summary of methods for solving triangles.
1. One side and two angles known. (SAA or ASA) 1. Find the third angle (A + B + C = 180o) 2. Find the remaining sides using sine rule 2. Two sides and any one angle known (SSA) 1. Use sine law to find an angle. 2. Find the third angle (A + B + C = 180o 3. Find the third side using sine rule. 3. Two sides and included angle known (SAS) 1. Find the third side using cosine rule. 2. Find the smaller of the remaining angles using sine rule . 3. Find the third angle (A + B + C) . 4. Three sides are known (SSS) 1. Find the largest angle using cosine rule. Use sine rule to find remaining two angles.

Solve the triangle with C = 28. 3o, b = 5. 71 cm, a = 4
Solve the triangle with C = 28.3o, b = 5.71 cm, a = 4.21 cm to two decimal places. A Two sides and the included angle are given. c b We use the cosine rule: B 28.30 c2 = a2 + b2 – 2 ab cos C C a = – 2 × 4.21 × 5.71 cos 28.3 = = cm Now we use the sine rule to find angles A and B = A = sin-1(0.7053) = 44.85o in the 1st quadrant We don’t need to solve for the second quadrant angle. B = 180 – ( o) = o

Solve the triangle with A = 67.3o, b = 37.9 m, c = 40.8 m.
Two sides and the included angle are given We use the cosine rule for the third side. a2 = b2 + c2 – 2 bc cos A = – 2 × 37.9 × 40.8 cos 67.3 = = m Now we use the sine rule to find angles B and C = C = sin-1(0.8613) = 59.5o B = 180 – ( ) = 53.2o

Solve the triangle with B = 168.2o, a = 15.1 m, c = 19.2 m.
Two sides and the included angle are given We use the cosine rule for the third side. b2 = a2 + c2 – 2 ac cos B = – 2 × 15.1 × 19.2 cos 168.2 = = m Now we use the sine rule to find angles A and C = A = sin-1(0.0906) = 5.2o C = 180 – ( o) = 6.6o

Solve the triangle with a = 3.0 m, b = 5.0 m, c = 6.0 m.
Three sides are known We find C, the largest angle opposite to the largest side c = 6.0 m c2 = a2 + b2 – 2 ab cos C 2 ab cos C = a2 + b2 – c2 = C = cos-1( ) = 94o Use sine rule to find one of the remaining angles. = A= sin-1(0.4989) = 30o B = 180 – (93.8o o) = 56o

Solve the triangle with a = 187 m, b = 214 m, c = 325 m.
Three sides are known We find C, the largest angle opposite to the largest side c = 325 m c2 = a2 + b2 – 2 ab cos C 2 ab cos C = a2 + b2 – c2 = C = cos-1( ) = 108.1o Use sine rule to find one of the remaining angles. = A= sin-1(0.5469) = 33.2o B = 180 – (108.1o o) = 38.7o

Since b = c, B = C (isosceles triangle) A Harbor
Two ships leave a harbor together, traveling on courses that make an angle 135o 40’. How far apart are they when they each travel 402 km? A is the harbor a B C AC = AB = 402 km c A = 135o 40’ b 135o 40’ Since b = c, B = C (isosceles triangle) A Harbor B + C = 180 – 135o 40’ = 44o 20’ B = (44o 20’)/2 = 22o 10’ We find side BC = a = 745 km

ABCD is a parallelogram. C a b A = C = 58o B = D = 122o d A
The sides of a parallelogram are 4.0 cm and 6.0 cm. One angle is 58o while another is 122o. Find the lengths of its diagonals. D ABCD is a parallelogram. C 122o a b A = C = 58o B = D = 122o 58o d A DB = a is a diagonal. B d AB = 6.0 cm AD = 4.0 cm Using cosine law in ABD BD2 = AD2 + AB2 – 2 × AD × AB cos A = – 2 × 4 × 6 cos 58o = 26.56 = 5.2 cm AC = is a diagonal. Using cosine law in ADB AC2 = AD2 + DC2 – 2 AD × DC cos D = – 2 × 4 × 6 cos 122 = – 2 × 4 × 6 cos 122 = 77.4 = 8.8 cm

Solve the triangle with C = 58.4o, b = 7.23 cm, c = 6.54 cm.
This is a case of SSA Use sine rule to find one of the missing angles. C a b B A c = B1 = sin-1(0.9416) = 70.3o in the 1st quadrant B2 = 180 – 70.3 = in the 2nd quadrant. A1 = 180 – (58.4o +70.3o) = 51.3o A2 = 180 – (58.4o o) = 11.9o = 5.99 cm = 1.58 cm

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