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AS-Level Maths: Core 2 for Edexcel

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.

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AS-Level Maths: Core 2 for Edexcel

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  1. AS-Level Maths:Core 2for 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

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

  3. The sine rule h h a b 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 a is the side opposite A and b is the side opposite B. A B D sin B = sin A = h = a sin B h = b sin A b sin A = a sin B So:

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

  5. The sine rule C b a A B c b c a sin A sin B sin C = = = = a b c sin B sin C sin A For any triangle ABC: or

  6. Using the sine rule to find side lengths B 39° 7 a = a sin 39° sin 118° 118° 7 sin 118° a = C A 7 cm sin 39° 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 = 9.82cm (to 2 d.p.)

  7. Using the sine rule to find angles C sin 46° sin B = 8 cm 8 6 6 cm 8 sin 46° sin B = 6 8 sin 46° 46° A B sin–1 B= 6 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: B = 73.56° (to 2 d.p.)

  8. Finding the second possible value C 8 cm 6 cm 6 cm 46° 46° A B B Suppose that in the last examplewe 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. 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.)

  9. Using the sine rule to solve triangles

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

  11. 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 a is the side opposite A and b is the side opposite B. A B x D c – x c is the side opposite C. If AD = x, then the length BD can be written as c – x.

  12. The cosine rule 1 2 cos A = x b Substituting and gives: 1 2 Using Pythagoras’ theorem in triangle ACD: C b2 = x2 +h2 b a h Also: A B x D c – x x = b cos A In triangle BCD: a2 = (c – x)2 + h2 a2 = c2 – 2cx+ x2 + h2 a2 = c2 – 2cx + x2 + h2 This is the cosine rule. a2 = c2 – 2cbcos A+ b2 a2 = b2 + c2 – 2bc cos A

  13. The cosine rule A c b b2 + c2 – a2 B C cos A = a 2bc For any triangle ABC: a2 = b2 + c2 – 2bc cos A or

  14. Using the cosine rule to find side lengths B a 4 cm 48° A C 7 cm 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. a2 = b2 + c2 – 2bc cos A a2 = 72 + 42 – (2× 7 × 4 × cos 48°) a2 = 27.53 (to 2 d.p.) a = 5.25 cm (to 2 d.p.)

  15. Using the cosine rule to find angles B b2 + c2 – a2 cos A = 2bc 8 cm 42 + 62 – 82 cos A = 6 cm 2 × 4 × 6 C A 4 cm 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. This is negative so A must be obtuse. cos A = –0.25 A = cos–1 –0.25 A = 104.48° (to 2 d.p.)

  16. Using the cosine rule to solve triangles

  17. The area of a triangle using ½ ab sinC • 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 Contents 17 of 40 © Boardworks Ltd 2005

  18. The area of a triangle A 4 cm 47° C B = sin 47° 7 cm h 4 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? We can find the height h using the sine ratio. h h = 4 sin 47° Area of triangle ABC = ½ × base × height = ½ × 7 × 4 sin 47° = 10.2 cm2(to 1 d.p.)

  19. The area of a triangle using ½ ab sin C 1 Area of triangle ABC = ab sin C 2 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 C a

  20. The area of a triangle using ½ ab sin C

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

  22. Measuring angles in degrees The system of using degrees to measure angles, where 1° is equal to of a full turn, is attributed to the ancient Babylonians. For example, of a full turn is equal to 160°. An angle is a measure of rotation. 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.

  23. Measuring angles in radians r r 1 rad O r So: 1 rad = 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. 2π rad = 360°

  24. Converting radians to degrees 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: We can convert radians to degrees using: 2π rad = 360° Radians are usually expressed as fractions or multiples of πso, for example:

  25. Converting degrees to radians To convert degrees to radians we multiply by . For example: 10 3 9 Sometimes angles are required to a given number of decimal places, rather than as multiples of π,for example: Note that when radians are written in terms of π the units rad or c are not normally needed.

  26. Converting between degrees and radians

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

  28. Using radians to measure arc length A r θ B O r 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. If the angle at the centre is 2 radians then the length of the arc is 2r. If the angle at the centre is 0.3 radians then the length of the arc is 0.3r. In general: Length of arc AB = θr where θis measured in radians. When θis measured in degrees the length of AB is

  29. Finding the area of a sector Again suppose an arc AB subtends an angle of θradians at the centre O of a circle. A So the area of the sector AOB is of the area of the full circle. r θ B O r  Area of sector AOB = Area of sector AOB = r2θ In general: We can also find the area of a sector using radians. The angle at the centre of a full circle is 2π radians. where θis measured in radians. When θis measured in degrees the area of AOB is

  30. 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. A B O 9 cm b) area of sector AOB = r2θ a) length of arc AB = θr = 6π cm = 27π cm2

  31. Finding the area of a segment A 45° B O 5 cm 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. Let’s call the area of sector AOB AS and the area of triangle AOB AT.

  32. Finding the area of a segment 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: where θis measured in radians.

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

  34. Solving equations using radians Solve 4 cos 2θ= 2 for . 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: 4 cos 2θ= 2 So: cos 2θ= 0.5 Changing the range to match the multiple angle: –π≤ 2θ≤ π If we now let x = 2θwe can solve cos x = 0.5 in the range –π≤ x ≤ π.

  35. Solving equations using radians The principal solution of cos x = 0.5 is if x = θ = This is the complete solution set in the range 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:

  36. Examination-style questions • 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 Contents 36 of 40 © Boardworks Ltd 2005

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

  38. Examination-style question 1 Area of triangle ABC = × 6 × 7 × sin 13.34° b) Area of triangle ABC = ac sin B where a = 6 cm, c = 7 cm and B = (180 – 50 – 116.66)° = 13.34° = 4.85 cm2 (to 2 d.p.)

  39. Examination-style question 2 A B 6 cm θ O 10 cm D C 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. = θradians. • Find an expression for the area of the shaded region in terms of θ. • Given that the shaded region is 25.6 cm2 find the value of θ. • Calculate the perimeter of the shaded region.

  40. Examination-style question 2 Area of sector BOD = × 62 × θ b) 32θ = 25.6 a) Area of sector AOC = × 102 × θ = 50θ = 18θ Area of shaded region = 50θ – 18θ = 32θ 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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