JR/2008 Trigonometry – non-right angled triangles Work up to now with basic trig. ratios has needed a right angled triangle for the ratios to be applied. This is also true for Pythagoras theorem. In many engineering examples it is useful to be able to obtain sides and angles of triangles which do not contain a right angle. To solve non-right angled triangles two rules are used: a.the sine rule b.the cosine rule

JR/2008 Trigonometry – non-right angled triangles Triangle Notation Before using the rules the correct notation for the triangle must be used. All angles are given uppercase letters (usually A, B, C) This is usually done in a clockwise sense All sides are given lowercase letters to match the opposite angle A B C c b a

JR/2008 a= b=c sin A sin B sin C Trigonometry – non-right angled triangles The Sine rule A B C c b a

JR/2008 Trigonometry – non-right angled triangles The Sine rule The sine rule can be applied to any triangle if: i.1 side length and 2 angles are known. If two angles are known the other is found knowing all angles add up to 180° Here side a and angles A and C are known a= b = c sin A sin B sin C A B C c b a

JR/2008 Trigonometry – non-right angled triangles The Sine rule If side c is required the rule is simply transposed Note – the unused terms are left out at this stage froma = b = c sin A sin B sin C We can use the part of the rule needed: a = c sin A sin C Which provides:a sin C = c allowing side c to be found sin A A B C c b a

JR/2008 Trigonometry – non-right angled triangles The Sine rule The sine rule can be applied to any triangle if: ii.2 side lengths and 1 non-included angle is known. Here sides a and b and angle A are known a= b = c sin A sin B sin C A B C c b a

JR/2008 Trigonometry – non-right angled triangles The Sine rule If Angle B is required the rule is simply transposed Note – with angle B known angle C can be obtained and the previous example used to obtain side c froma = b = c sin A sin B sin C We can use the part of the rule needed:a = b sin A sin B Which provides:sin B = b sin A a A B C c b a

JR/2008 a 2 = b 2 + c 2 - 2bc Cos A orb 2 = a 2 + c 2 - 2ac Cos B orc 2 = a 2 + b 2 - 2ab Cos C Trigonometry – non-right angled triangles The Cosine rule A B C c b a

JR/2008 Trigonometry – non-right angled triangles The Cosine rule The cosine rule can be applied to any triangle if: i.2 side lengths and 1 included angle are known. Here sides b and c and angle A are known a 2 = b 2 + c 2 - 2bc Cos A To find side length a = A B C c b a

JR/2008 Trigonometry – non-right angled triangles The Cosine rule The cosine rule can be applied to any triangle if: ii.3 side lengths are known. Here sides a, b and c are known a 2 = b 2 + c 2 - 2bc Cos A To find Angle A = A B C c b a

Due to the complexity of the cosine rule it is usually only used once when solutions to non right angled triangles are needed. Once the cosine rule has been used to obtain an angle or a side the sine rule is then selected to finish solving the triangle fully. Typical example on the use of sin and cosine rules. Find all of the angles in the triangle shown below when: Side a = 68 mm Side b = 74 mm Side c = 47 mm JR/2008 Trigonometry – non-right angled triangles The sin/cosine rule A B C c b a

The first step is to realise that the triangle does not contain a right angle and decide that the cosine rule can be used to obtain one of the angles To find angle A use a 2 = b 2 + c 2 - 2bc Cos A Transpose for A A = when: a = 68, b = 74, c = 47 Angle A = Angle A = cos -1 (0.4401) JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule A B C As the cosine is +ve the angle could be in the 1 st or 4 th quadrant. The result could be Angle A = 63.9° or 360° – 63.9° = 296.1°. Inspection shows the former to be correct as angles in a triangle add up to 180°

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule As angle A is known it is now possible to use the sine rule to find angle C thus when: a = 68, c = 47, A = 63.9° Angle C = Angle C = sin -1 (0.6207) Angle C = 38.37° (38.4°) Note: Previous work on angles in a quadrant will show two possible values of angle C with a sine of (1 st or 2 nd quadrant) Angle C could also have been 180° – 38.4° = 141.6° Inspection of the triangle shows this result to be false. A B C

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule As two angles are known it is now possible to find angle B If angle C is assumed as 38.4° then Angle B = 180° - (38.4° °) = 77.7° To confirm the correct solution find Angle B using the sine rule: Angle B = = sin Angle B = 77.7° As sin is +ve in 1 st & 2 nd quadrant solution could also be: 180 ° ° = 257.7° not sensible A B C

JR/2008 Trigonometry – non-right angled triangles The sin/cosine rule A more complex solution is found when one of the angles in the given triangle > 90° Typical example on the use of sin and cosine rules with angle B > 90° Find all of the angles in the triangle shown below when: Side a = 52 mm Side b = 65 mm Side c = 36 mm A B C c b a

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule The first step is to realise that the triangle does not contain a right angle and decide that the cosine rule can be used to obtain one of the angles To find angle A use a 2 = b 2 + c 2 - 2bc Cos A Transpose for A A = when: a = 52, b = 65, c = 36 Angle A = Angle A = cos -1 (0.6019) Angle A = 53° Note: Angle could also have been 360° – 53° = 307° Inspection shows this not to be a sensible result A B C

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule As angle A is known it is now possible to use the sine rule to find angle B thus when: a = 52, b = 65, A = 53° Angle B = Angle B = sin -1 (0.9983) Angle B = 86.6° Note: Previous work on angles in a quadrant will show two possible values of angle B with a sine of (1 st or 2 nd quadrant) Angle B could also have been 180° – 86.6° = 93.4° A result this close should generate a complete re-check of all calculations to ensure the correct value is used. A B C °

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule As two angles are known it is now possible to find angle C If angle B is assumed as 86.65° then Angle C = 180° - (86.65° + 53°) = 40.4° If angle B is assumed as 93.35° then Angle C = 180° - (93.35° + 53°) = 33.6° To confirm the correct solution find Angle C using the sine rule: Angle C == 33.6° As sin is +ve in 1 st & 2 nd quadrant solution could also be: 180 ° ° = 146.4° inspection shows the former to be correct A B C ° ?

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule As angle C is found to be 33.6° then angle B must be 93.4° A B C ° 93.4° 33.6°

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule A practical use of the sin rule within a circle The circle diameter can be found from the sine rule. This can be useful when looking to find the pitch circle diameter (pcd) for a ring of holes A B C b a c Diameter D

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule If centre to centre distance a = mm and centre distance b = mm find the pcd of the three holes.Angle A = 33° side a = thus Dia D = mm A B C b a c pcd 33°

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule Examples – fill in blank spaces A°A°B°B°C°C°abc 75 °34 °102mm 71 °61 °91mm 116 °18 °170mm 90 °35 °50mm 39 °17mm15mm 115°4’11°17’516.2mm 105°4’9.217m7.152m 60°9m11m 56 °10m14m 312mm527.3mm700mm A B C

JR/2008 Trigonometry – non-right angled triangles The sine/cosine rule Examples - solutions A°A°B°B°C°C°abc 75 °34 °71°102mm59.1mm99.9mm 48°71 °61 °71.5mm91mm84.2mm 116 °46°18 °170mm136mm58.4mm 55°90 °35 °50mm61mm35mm 45.5°/ ° 95.5°/6.5° 17mm15mm 23.7/2.7mm 53°39’115°4’11°17’2124mm2390mm516.2mm 105°4’26°25’48°31’9.217m4.247m7.152m 50°11’69°49’60°9m11m10.2m 79°18’44°42’56 °10m14m11.8m 24°42’44°54’110°24’312mm527.3mm700mm A B C