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© Boardworks Ltd 2005 1 of 40 © Boardworks Ltd 2005 1 of 40 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.

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© Boardworks Ltd 2005 2 of 40 Contents © Boardworks Ltd 2005 2 of 40 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 The sine rule

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© Boardworks Ltd 2005 3 of 40 The sine rule Consider any triangle ABC: C AB ba If we drop a perpendicular h from C to AB, we can divide the triangle into two right-angled triangles; ACD and BDC. D sin A = h a is the side opposite A and b is the side opposite B. h b h = b sin A sin B = h a h = a sin B b sin A = a sin B So:

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© Boardworks Ltd 2005 4 of 40 The sine rule b sin A = a sin B Dividing both sides of the equation by sin A and then by sin B gives: b sin B = a sin A If we had dropped a perpendicular from A to BC we would have found that: b sin C = c sin B Rearranging: b sin B = c sin C

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© Boardworks Ltd 2005 5 of 40 The sine rule For any triangle ABC: C AB b c a a sin A = b sin B = c sin C or sin A sin B sin C a = b = c

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© Boardworks Ltd 2005 6 of 40 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. Using the sine rule: a sin 118° = 7 sin 39° a = 7 sin 118° sin 39° a = 9.82 cm (to 2 d.p.) a 7 cm 118° 39° A B C

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© Boardworks Ltd 2005 7 of 40 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. Using the sine rule: sin B 8 = 6 sin 46° sin B = 8 sin 46° 6 B = 73.56° (to 2 d.p.) sin –1 B =B = 8 sin 46° 6 6 cm 46° B 8 cm A C

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© Boardworks Ltd 2005 8 of 40 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°. 6 cm 46° B 8 cm A C There is a second possible value for the angle at B. Instead of this triangle … … we could have this triangle. Remember: sin θ = sin (180° – θ ) So for every acute solution, there is a corresponding obtuse solution. B = 73.56° (to 2 d.p.) or B = 180° – 73.56° = 106.44° (to 2 d.p.) 46° 6 cm B

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© Boardworks Ltd 2005 9 of 40 Using the sine rule to solve triangles

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© Boardworks Ltd 2005 10 of 40 Contents © Boardworks Ltd 2005 10 of 40 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 The cosine rule

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© Boardworks Ltd 2005 11 of 40 The cosine rule Consider any triangle ABC: C AB b a If we drop a perpendicular h from C to AB, we can divide the triangle into two right-angled triangles; ACD and BDC. D h 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. c – x x

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© Boardworks Ltd 2005 12 of 40 The cosine rule C AB b a Using Pythagoras’ theorem in triangle ACD: D h b 2 = x 2 + h 2 c – x x Also: cos A = x b In triangle BCD: x = b cos A a 2 = ( c – x ) 2 + h 2 a 2 = c 2 – 2 cx + x 2 + h 2 1 2 Substituting and gives: 12 a 2 = c 2 – 2 cb cos A + b 2 a 2 = b 2 + c 2 – 2 bc cos A a 2 = c 2 – 2 cx + x 2 + h 2 This is the cosine rule.

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© Boardworks Ltd 2005 13 of 40 The cosine rule For any triangle ABC: A BC c a b a 2 = b 2 + c 2 – 2 bc cos A or cos A = b 2 + c 2 – a 2 2 bc

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© Boardworks Ltd 2005 14 of 40 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 C A 7 cm 4 cm 48° a a 2 = b 2 + c 2 – 2 bc cos A a 2 = 7 2 + 4 2 – (2 × 7 × 4 × cos 48°) a 2 = 27.53 (to 2 d.p.) a = 5.25 cm (to 2 d.p.)

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© Boardworks Ltd 2005 15 of 40 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. 4 cm 8 cm 6 cm A B C cos A = b 2 + c 2 – a 2 2 bc cos A = 4 2 + 6 2 – 8 2 2 × 4 × 6 cos A = –0.25 A = cos –1 –0.25 A = 104.48° (to 2 d.p.) This is negative so A must be obtuse.

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© Boardworks Ltd 2005 16 of 40 Using the cosine rule to solve triangles

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© Boardworks Ltd 2005 17 of 40 Contents © Boardworks Ltd 2005 17 of 40 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 The area of a triangle using ½ ab sin C

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© Boardworks Ltd 2005 18 of 40 The area of a triangle 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 B C 7 cm 4 cm 47° h We can find the height h using the sine ratio. h 4 = sin 47° h = 4 sin 47° Area of triangle ABC = ½ × base × height = ½ × 7 × 4 sin 47° = 10.2 cm 2 (to 1 d.p.) The area of a triangle is given by ½ × base × height.

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© Boardworks Ltd 2005 19 of 40 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 BC c a b Area of triangle ABC = ab sin C 1 2

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© Boardworks Ltd 2005 20 of 40 The area of a triangle using ½ ab sin C

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© Boardworks Ltd 2005 21 of 40 Contents © Boardworks Ltd 2005 21 of 40 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 Degrees and radians

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© Boardworks Ltd 2005 22 of 40 Measuring angles in degrees The use of the number 360 is thought to originate from the approximate number of days in a year. The system of using degrees to measure angles, where 1° is equal to of a full turn, is attributed to the ancient Babylonians. 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. An angle is a measure of rotation. For example, of a full turn is equal to 160°.

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© Boardworks Ltd 2005 23 of 40 Measuring angles in radians A full turn is divided into 2 π radians. One radian is therefore equal to the angle subtended by an arc of length r. In many mathematical and scientific applications, particularly in calculus, it is more appropriate to measure angles in radians. Remember that the circumference of a circle of radius r is equal to 2 πr. 1 radian can be written as 1 rad or 1 c. 2 π rad = 360° So: 1 rad = r r r 1 rad O

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© Boardworks Ltd 2005 24 of 40 Converting radians to degrees We can convert radians to degrees using: Radians are usually expressed as fractions or multiples of π so, for example: 2 π rad = 360° Or: π rad = 180° 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 For example:

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© Boardworks Ltd 2005 25 of 40 Converting degrees to radians To convert degrees to radians we multiply by. For example: Sometimes angles are required to a given number of decimal places, rather than as multiples of π, for example: 39 10 Note that when radians are written in terms of π the units rad or c are not normally needed.

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© Boardworks Ltd 2005 26 of 40 Converting between degrees and radians

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© Boardworks Ltd 2005 27 of 40 Contents © Boardworks Ltd 2005 27 of 40 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 Arc length and sector area

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© Boardworks Ltd 2005 28 of 40 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. r r θ O A B where θ is measured in radians. If the angle at the centre is 1 radian then the length of the arc is r. If the angle at the centre is 2 radians then the length of the arc is 2 r. If the angle at the centre is 0.3 radians then the length of the arc is 0.3 r. In general: Length of arc AB = θr When θ is measured in degrees the length of AB is

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© Boardworks Ltd 2005 29 of 40 Finding the area of a sector We can also find the area of a sector using radians. where θ is measured in radians. r r θ O A B 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 is 2 π radians. When θ is measured in degrees the area of AOB is In general: Area of sector AOB = r 2 θ Area of sector AOB = So the area of the sector AOB is of the area of the full circle.

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© Boardworks Ltd 2005 30 of 40 Finding chord length and sector area A chord AB subtends an angle of 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. 9 cm O A B a) length of arc AB = θr = 6 π cm = 27 π cm 2 b) area of sector AOB = r 2 θ

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© Boardworks Ltd 2005 31 of 40 Finding the area of a segment 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. 5 cm 45° O A B 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: Let’s call the area of sector AOB A S and the area of triangle AOB A T.

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© Boardworks Ltd 2005 32 of 40 Finding the area of a segment Now: Area of the minor segment = A S – A T = 9.8174… – 8.8388… = 0.979 cm 2 (to 3 sig. figs.) In general, the area of a segment of a circle of radius r is: where θ is measured in radians.

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© Boardworks Ltd 2005 33 of 40 Contents © Boardworks Ltd 2005 33 of 40 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 Solving equations using radians

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© Boardworks Ltd 2005 34 of 40 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 2 θ = 2 for. 4 cos 2 θ = 2 So: cos 2 θ = 0.5 Changing the range to match the multiple angle: – π ≤ 2 θ ≤ π If we now let x = 2 θ we can solve cos x = 0.5 in the range – π ≤ x ≤ π.

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© Boardworks Ltd 2005 35 of 40 Solving equations using radians The principal solution of cos x = 0.5 is 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 = 2 θ, so: θ = This is the complete solution set in the range if x =

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© Boardworks Ltd 2005 36 of 40 Contents © Boardworks Ltd 2005 36 of 40 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

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© Boardworks Ltd 2005 37 of 40 Examination-style question 1 In the triangle ABC, AB = 7 cm, BC = 6 cm and = 50°. a)Calculate the two possible sizes of in degrees to two decimal places. b)Given that is obtuse, calculate the area of triangle ABC to two decimal places. a) Using the sine rule: sin C 7 = 6 sin 50° sin B = 7 sin 50° 6 B = 63.34° or 116.66° (to 2 d.p.) sin –1 B =B = 7 sin 50° 6

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© Boardworks Ltd 2005 38 of 40 Examination-style question 1 b) Area of triangle ABC = ac sin B where a = 6 cm, c = 7 cm and B = (180 – 50 – 116.66)° = 13.34° Area of triangle ABC = × 6 × 7 × sin 13.34° = 4.85 cm 2 (to 2 d.p.)

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© Boardworks Ltd 2005 39 of 40 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. = θ radians. a)Find an expression for the area of the shaded region in terms of θ. b)Given that the shaded region is 25.6 cm 2 find the value of θ. c)Calculate the perimeter of the shaded region. O θ A B C D 6 cm 10 cm

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© Boardworks Ltd 2005 40 of 40 Examination-style question 2 a) Area of sector AOC = × 10 2 × θ Area of sector BOD = × 6 2 × θ Area of shaded region = 50 θ – 18 θ = 32 θ = 50 θ = 18 θ b)32 θ = 25.6 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

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