Presentation on theme: "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."— Presentation transcript:
1 The Law of CosinesThe 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:Caba2 = b2 + c2 – 2 bc cos ABcAIf a and b are two sides and C the included angle:c2 = a2 + b2 – 2 ab cos CIf a and c are two sides and B the included angle:b2 = a2 + c2 – 2 ac cos B
2 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 rule2.Two sides and any one angle known (SSA)1. Use sine law to find an angle.2. Find the third angle (A + B + C = 180o3. 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 twoangles.
3 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.ATwo sides and the included angle are given.cbWe use the cosine rule:B28.30c2 = a2 + b2 – 2 ab cos CCa= – 2 × 4.21 × 5.71 cos 28.3== cmNow we use the sine rule to find angles A and B=A = sin-1(0.7053) = 44.85o in the 1st quadrantWe don’t need to solve for the second quadrant angle.B = 180 – ( o) = o
4 Solve the triangle with A = 67.3o, b = 37.9 m, c = 40.8 m. Two sides and the included angle are givenWe use the cosine rule for the third side.a2 = b2 + c2 – 2 bc cos A= – 2 × 37.9 × 40.8 cos 67.3== mNow we use the sine rule to find angles B and C=C = sin-1(0.8613) = 59.5oB = 180 – ( ) = 53.2o
5 Solve the triangle with B = 168.2o, a = 15.1 m, c = 19.2 m. Two sides and the included angle are givenWe use the cosine rule for the third side.b2 = a2 + c2 – 2 ac cos B= – 2 × 15.1 × 19.2 cos 168.2== mNow we use the sine rule to find angles A and C=A = sin-1(0.0906) = 5.2oC = 180 – ( o) = 6.6o
6 Solve the triangle with a = 3.0 m, b = 5.0 m, c = 6.0 m. Three sides are knownWe find C, the largest angle opposite to the largest side c = 6.0 mc2 = a2 + b2 – 2 ab cos C2 ab cos C = a2 + b2 – c2=C = cos-1( )= 94oUse sine rule to find one of the remaining angles.=A= sin-1(0.4989)= 30oB = 180 – (93.8o o)= 56o
7 Solve the triangle with a = 187 m, b = 214 m, c = 325 m. Three sides are knownWe find C, the largest angle opposite to the largest side c = 325 mc2 = a2 + b2 – 2 ab cos C2 ab cos C = a2 + b2 – c2=C = cos-1( )= 108.1oUse sine rule to find one of the remaining angles.=A= sin-1(0.5469)= 33.2oB = 180 – (108.1o o)= 38.7o
8 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 harboraBCAC = AB = 402 kmcA = 135o 40’b135o 40’Since b = c, B = C (isosceles triangle)A HarborB + C = 180 – 135o 40’= 44o 20’B = (44o 20’)/2= 22o 10’We find side BC = a= 745 km
9 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.DABCD is a parallelogram.C122oabA = C = 58oB = D = 122o58odADB = a is a diagonal.BdAB = 6.0 cmAD = 4.0 cmUsing cosine law in ABDBD2 = AD2 + AB2 – 2 × AD × AB cos A= – 2 × 4 × 6 cos 58o= 26.56= 5.2 cmAC = is a diagonal.Using cosine law in ADBAC2 = AD2 + DC2 – 2 AD × DC cos D= – 2 × 4 × 6 cos 122= – 2 × 4 × 6 cos 122= 77.4= 8.8 cm
10 Solve the triangle with C = 58.4o, b = 7.23 cm, c = 6.54 cm. This is a case of SSAUse sine rule to find one of the missing angles.CabBAc=B1 = sin-1(0.9416) = 70.3o in the 1st quadrantB2 = 180 – 70.3 = in the 2nd quadrant.A1 = 180 – (58.4o +70.3o) = 51.3oA2 = 180 – (58.4o o) = 11.9o= 5.99 cm= 1.58 cm
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