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AS-Level Maths: Core 2 for Edexcel C2.5 Trigonometry 2 This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 40 © Boardworks Ltd 2005 The sine rule The sine rule Contents The cosine rule The area of a triangle using ½ ab sinC Degrees and radians Arc length and sector area Solving equations using radians Examination-style questions 2 of 40 © Boardworks Ltd 2005 The sine rule Consider any triangle ABC: If we drop a perpendicular h from C to AB, we can divide the triangle into two right-angled triangles; ACD and BDC. C b A h D h b h = b sin A sin A = So: 3 of 40 a B a is the side opposite A and b is the side opposite B. h a h = a sin B sin B = b sin A = a sin B © Boardworks Ltd 2005 The sine rule b sin A = a sin B Dividing both sides of the equation by sin A and then by sin B gives: a b = sin A sin B If we had dropped a perpendicular from A to BC we would have found that: b sin C = c sin B Rearranging: c b = sin C sin B 4 of 40 © Boardworks Ltd 2005 The sine rule For any triangle ABC: C b A b c a = = sin B sin C sin A 5 of 40 a c or B sin A sin B sin C = = a b c © Boardworks Ltd 2005 Using the sine rule to find side lengths If we are given two angles in a triangle and the length of a side opposite one of the angles, we can use the sine rule to find the length of the side opposite the other angle. For example: Find the length of side a. B 39° a Using the sine rule: 7 a = sin 39° sin 118° 118° 7 cm C A 7 sin 118° a = sin 39° a = 9.82 cm (to 2 d.p.) 6 of 40 © Boardworks Ltd 2005 Using the sine rule to find angles If we are given two side lengths in a triangle and the angle opposite one of the given sides, we can use the sine rule to find the angle opposite the other given side. For example: Find the angle at B. C 8 cm A 46° 7 of 40 6 cm B Using the sine rule: sin 46° sin B = 8 6 8 sin 46° sin B = 6 8 sin 46° –1 B = sin 6 B = 73.56° (to 2 d.p.) © Boardworks Ltd 2005 Finding the second possible value Suppose that in the last example we had not been given a diagram but had only been told that AC = 8 cm, CB = 6 cm and that the angle at A = 46°. There is a second possible value for the angle at B. Instead of this triangle … … we could have this triangle. C Remember: sin θ = sin (180° – θ) So for every acute solution, there is a corresponding obtuse solution. 8 cm 6 cm B = 73.56° (to 2 d.p.) 6 cm or 46° B = 180° – 73.56° A B B = 106.44° (to 2 d.p.) 8 of 40 © Boardworks Ltd 2005 Using the sine rule to solve triangles 9 of 40 © Boardworks Ltd 2005 The cosine rule Contents The sine rule The cosine rule The area of a triangle using ½ ab sinC Degrees and radians Arc length and sector area Solving equations using radians Examination-style questions 10 of 40 © Boardworks Ltd 2005 The cosine rule Consider any triangle ABC: If we drop a perpendicular h from C to AB, we can divide the triangle into two right-angled triangles; ACD and BDC. C b A h x a D c–x B a is the side opposite A and b is the side opposite B. c is the side opposite C. If AD = x, then the length BD can be written as c – x. 11 of 40 © Boardworks Ltd 2005 The cosine rule Using Pythagoras’ theorem in triangle ACD: C b A a h D c–x x In triangle BCD: Substituting 1 and b2 = x2 + h2 1 Also: B x cos A = b x = b cos A 2 a2 = (c – x)2 + h2 a2 = c2 – 2cx + x2 + h2 gives: a2 = c2 – 2cb cos A + b2 2 This is the cosine rule. a2 = b2 + c2 – 2bc cos A 12 of 40 © Boardworks Ltd 2005 The cosine rule For any triangle ABC: a2 = b2 + c2 – 2bc cos A A c B 13 of 40 or b a C cos A = b2 + c2 – a2 2bc © Boardworks Ltd 2005 Using the cosine rule to find side lengths If we are given the length of two sides in a triangle and the size of the angle between them, we can use the cosine rule to find the length of the other side. For example: Find the length of side a. B a a2 = b2 + c2 – 2bc cos A 4 cm 48° C 7 cm a2 = 72 + 42 – (2 × 7 × 4 × cos 48°) A a2 = 27.53 (to 2 d.p.) a = 5.25 cm (to 2 d.p.) 14 of 40 © Boardworks Ltd 2005 Using the cosine rule to find angles If we are given the lengths of all three sides in a triangle, we can use the cosine rule to find the size of any one of the angles in the triangle. For example: Find the size of the angle at A. B 8 cm 6 cm cos A = b2 + c2 – a2 2bc 2 + 62 – 82 4 cos A = 2×4×6 cos A = –0.25 C 4 cm A This is negative so A must be obtuse. A = cos–1 –0.25 A = 104.48° (to 2 d.p.) 15 of 40 © Boardworks Ltd 2005 Using the cosine rule to solve triangles 16 of 40 © Boardworks Ltd 2005 The area of a triangle using ½ ab sinC Contents The sine rule The cosine rule The area of a triangle using ½ ab sin C Degrees and radians Arc length and sector area Solving equations using radians Examination-style questions 17 of 40 © Boardworks Ltd 2005 The area of a triangle The area of a triangle is given by ½ × base × height. Suppose that instead of the height of a triangle, we are given the base, one of the sides and the included angle. For example: What is the area of triangle ABC? A h 4 cm 47° B 7 cm We can find the height h using the sine ratio. C h = sin 47° 4 h = 4 sin 47° Area of triangle ABC = ½ × base × height = ½ × 7 × 4 sin 47° = 10.2 cm2 (to 1 d.p.) 18 of 40 © Boardworks Ltd 2005 The area of a triangle using ½ ab sin C In general, the area of a triangle is equal to half the product of two of the sides and the sine of the included angle. A c B b a Area of triangle ABC = 19 of 40 C 1 ab sin C 2 © Boardworks Ltd 2005 The area of a triangle using ½ ab sin C 20 of 40 © Boardworks Ltd 2005 Degrees and radians Contents The sine rule The cosine rule The area of a triangle using ½ ab sin C Degrees and radians Arc length and sector area Solving equations using radians Examination-style questions 21 of 40 © Boardworks Ltd 2005 Measuring angles in degrees An angle is a measure of rotation. The system of using degrees to measure angles, where 1° is equal to 3 61 0 of a full turn, is attributed to the ancient Babylonians. The use of the number 360 is thought to originate from the approximate number of days in a year. 360 is also a number that has a high number of factors and so many fractions of a full turn can be written as a whole number of degrees. For example, 94 of a full turn is equal to 160°. 22 of 40 © Boardworks Ltd 2005 Measuring angles in radians In many mathematical and scientific applications, particularly in calculus, it is more appropriate to measure angles in radians. A full turn is divided into 2π radians. Remember that the circumference of a circle of radius r is equal to 2πr. One radian is therefore equal to the angle subtended by an arc of length r. 1 radian can be written as 1 rad or 1c. r O 23 of 40 r 2π rad = 360° 1 rad r So: 1 rad = 360 2 57.3 © Boardworks Ltd 2005 Converting radians to degrees We can convert radians to degrees using: 2π rad = 360° π rad = 180° Or: Radians are usually expressed as fractions or multiples of π so, for example: ra d = 2 180 5 o 2 = 90 o 6 ra d = 5 × 180 o = 150 o 6 If the angle is not given in terms of π, when using a calculator for example, it can be converted to degrees by multiplying by 180 . For example: 2 .5 ra d = 2 .5 × 24 of 40 180 = 1 4 3 .2 4 (to 2 d .p .) © Boardworks Ltd 2005 Converting degrees to radians To convert degrees to radians we multiply by . 180 For example: o 60 = 60 × = 180 3 o ra d 10 200 = 200 × 3 180 = 9 1 0 ra d 9 Sometimes angles are required to a given number of decimal places, rather than as multiples of π, for example: o 37 = 37 × 180 = 0 .6 4 6 c (to 3 d .p .) Note that when radians are written in terms of π the units rad or c are not normally needed. 25 of 40 © Boardworks Ltd 2005 Converting between degrees and radians 26 of 40 © Boardworks Ltd 2005 Arc length and sector area Contents The sine rule The cosine rule The area of a triangle using ½ ab sin C Degrees and radians Arc length and sector area Solving equations using radians Examination-style questions 27 of 40 © Boardworks Ltd 2005 Using radians to measure arc length Suppose an arc AB of a circle of radius r subtends an angle of θ radians at the centre. If the angle at the centre is 1 radian then the length of the arc is r. A If the angle at the centre is 2 radians r then the length of the arc is 2r. θ B If the angle at the centre is 0.3 radians r O then the length of the arc is 0.3r. In general: Length of arc AB = θr where θ is measured in radians. × 2 r When θ is measured in degrees the length of AB is 360 28 of 40 © Boardworks Ltd 2005 Finding the area of a sector We can also find the area of a sector using radians. Again suppose an arc AB subtends an angle of θ radians at the centre O of a circle. The angle at the centre of a full circle A is 2π radians. r θ O r B So the area of the sector AOB is of the area of the full circle. Area of sector AOB = 2 2 ×r 2 In general: Area of sector AOB = 21 r2θ where θ is measured in radians. 2 ×r When θ is measured in degrees the area of AOB is 360 29 of 40 © Boardworks Ltd 2005 Finding chord length and sector area A chord AB subtends an angle of 23 radians at the centre O of a circle of radius 9 cm. Find in terms of π: a) the length of the arc AB. b) the area of the sector AOB. a) length of arc AB = θr A = 2 3 O 2 ×9 3 9 cm B = 6π cm b) area of sector AOB = 21 r2θ = 1 2 ×9 × 2 2 3 = 27π cm2 30 of 40 © Boardworks Ltd 2005 Finding the area of a segment The formula for the area of a sector can be combined with the formula for the area of a triangle to find the area of a segment. For example: A chord AB divides a circle of radius 5 cm into two segments. If AB subtends an angle of 45° at the centre of the circle, find the area of the minor segment to 3 significant figures. o 45 = A O 45° 5 cm ra d ia n s 4 B Let’s call the area of sector AOB AS and the area of triangle AOB AT. AS = = 1 2 1 2 r 2 2 ×5 × 4 = 9.8174... cm 31 of 40 2 © Boardworks Ltd 2005 Finding the area of a segment AT = = 1 2 1 2 r sin 2 2 × 5 × sin 4 = 8.8388... cm 2 Now: Area of the minor segment = AS – AT = 9.8174… – 8.8388… = 0.979 cm2 (to 3 sig. figs.) In general, the area of a segment of a circle of radius r is: A= 1 2 r ( sin ) 2 where θ is measured in radians. 32 of 40 © Boardworks Ltd 2005 Solving equations using radians Contents The sine rule The cosine rule The area of a triangle using ½ ab sin C Degrees and radians Arc length and sector area Solving equations using radians Examination-style questions 33 of 40 © Boardworks Ltd 2005 Solving equations using radians If the range for the solution set of a trigonometric equation is given in radians then the solution must also be given in radians. For example: Solve 4 cos 2θ = 2 for So: 2 2 . 4 cos 2θ = 2 cos 2θ = 0.5 Changing the range to match the multiple angle: –π ≤ 2θ ≤ π If we now let x = 2θ we can solve cos x = 0.5 in the range –π ≤ x ≤ π. 34 of 40 © Boardworks Ltd 2005 Solving equations using radians The principal solution of cos x = 0.5 is . 3 Remember that cos is an even function and so, in general, cos (–θ) = cos θ. the second solution for x in the range –π ≤ x ≤ π is x = But x = 2θ, so: if x = θ= , 3 3 6 , . 3 6 This is the complete solution set in the range 35 of 40 2 . 2 © Boardworks Ltd 2005 Examination-style questions Contents The sine rule The cosine rule The area of a triangle using ½ ab sin C Degrees and radians Arc length and sector area Solving equations using radians Examination-style questions 36 of 40 © Boardworks Ltd 2005 Examination-style question 1 In the triangle ABC, AB = 7 cm, BC = 6 cm and B A C = 50°. a) Calculate the two possible sizes of A C B in degrees to two decimal places. b) Given that A C B is obtuse, calculate the area of triangle ABC to two decimal places. a) Using the sine rule: sin 50° sin C = 7 6 7 sin 50° sin B = 6 7 sin 50° –1 B = sin 6 B = 63.34° or 116.66° (to 2 d.p.) 37 of 40 © Boardworks Ltd 2005 Examination-style question 1 b) Area of triangle ABC = 21 ac sin B where a = 6 cm, c = 7 cm and B = (180 – 50 – 116.66)° = 13.34° Area of triangle ABC = 21 × 6 × 7 × sin 13.34° = 4.85 cm2 (to 2 d.p.) 38 of 40 © Boardworks Ltd 2005 Examination-style question 2 In the following diagram AC is an arc of a circle with centre O and radius 10 cm and BD is an arc of a circle with centre O and radius 6 cm. AOD = θ radians. A a) Find an expression for the area of the shaded region in terms of θ. B 6 cm θ O 10 cm D C 39 of 40 b) Given that the shaded region is 25.6 cm2 find the value of θ. c) Calculate the perimeter of the shaded region. © Boardworks Ltd 2005 Examination-style question 2 a) Area of sector AOC = 21 × 102 × θ = 50θ Area of sector BOD = 2 × 62 × θ = 18θ 1 Area of shaded region = 50θ – 18θ = 32θ 32θ = 25.6 b) θ = 25.6 ÷ 32 θ = 0.8 radians c) Perimeter of the shaded region = length of arc AC + length of arc BD + AB + CD = (10 × 0.8) + (6 × 0.8) + 8 = 20.8 cm 40 of 40 © Boardworks Ltd 2005