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AS-Level Maths: Core 2 for Edexcel

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1 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

2 Contents 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 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 We call the perpendicular h for height (not h for hypotenuse). h b h a sin A = sin B = h = b sin A h = a sin B So: b sin A = a sin B

4 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

5 The sine rule For any triangle ABC: C A B b c a a sin A = b sin B c
We can use the first form of the equation to find side lengths and the second form of the equation to find angles. a sin A = b sin B c sin C sin A sin B sin C a = b c or

6 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. a 7 cm 118° 39° A B C Using the sine rule: a sin 118° = 7 sin 39° When trying to find a side length it is easier to use the sine rule equation in the form a/sin A = b/sin B . Encourage students to wait until the last step in the equation to evaluate the sines of the required angles. This avoids errors in rounding. a = 7 sin 118° sin 39° a = cm (to 2 d.p.)

7 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. 6 cm 46° B 8 cm A C Using the sine rule: sin B 8 = 6 sin 46° When trying to find an angle it is easier to use the sine rule equation in the form sin A/a = sin B/b. sin B = 8 sin 46° 6 sin–1 B = 8 sin 46° 6 B = 73.56° (to 2 d.p.)

8 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. 6 cm 46° B 8 cm A C Remember: sin θ = sin (180° – θ) 46° 6 cm B So for every acute solution, there is a corresponding obtuse solution. Remind students that the sine of angles in the second quadrant (that is, angles between 90° and 180°) are positive. This means that for every angle between 0° and 90° there is another angle between 90° and 180° that has the same sine. This angle is found by subtracting the associated acute angle from 180°. Ask students to imagine constructing the given triangle using a ruler and compasses (or ask them to do this as a practical). If the compass needle is placed at C and opened to 6 cm, there are two places that it can cross the line AB. These two points give the two possible triangles. When there are two possible angles it is often called “the ambiguous case”. B = 73.56° (to 2 d.p.) or B = 180° – 73.56° = ° (to 2 d.p.)

9 Using the sine rule to solve triangles
Be aware that the lengths of the sides and the angles have been rounded. Show the side lengths and angle sizes to a greater degree of accuracy to overcome this. Reveal an angle and the side opposite it. Reveal one more angle and ask students to find the side opposite it by using the sine rule. Alternatively, reveal an angle, the side opposite it and one more side and ask students to find the angle opposite it. Generate new examples by modifying the shape of the triangle.

10 Contents 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 B a is the side opposite A and b is the side opposite B. x D c – x We call the perpendicular h for height (not h for hypotenuse). c is the side opposite C. If AD = x, then the length BD can be written as c – x.

12 The cosine rule Using Pythagoras’ theorem in triangle ACD: C
b2 = x2 + h2 1 b a h Also: cos A = x b A B x D c – x x = b cos A 2 In triangle BCD: a2 = (c – x)2 + h2 We call the perpendicular h for height (not h for hypotenuse). To derive the cosine rule we use both Pythagoras’ theorem and the cosine ratio. a2 = c2 – 2cx + x2 + h2 a2 = c2 – 2cx + x2 + h2 Substituting and gives: 1 2 This is the cosine rule. a2 = c2 – 2cb cos A + b2 a2 = b2 + c2 – 2bc cos A

13 The cosine rule For any triangle ABC: A B C c a b
a2 = b2 + c2 – 2bc cos A or cos A = b2 + c2 – a2 2bc We can use the first form of the equation to find side lengths and the second form of the equation to find angles.

14 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 a2 = b2 + c2 – 2bc cos A The angle between two sides is often called the included angle. We can express the cosine rule as: “the square of the unknown side is equal to the sum of the squares of the other two sides minus 2 times the product of the other two sides and the cosine of the included angle”. Warn students not to forget to find the square root. The answer should look sensible considering the other lengths. Advise students to keep the value for a2 on their calculator displays. They should square root this value rather than the rounded value that has been written down. This will avoid possible errors in rounding. Point out that if we are given the lengths of two sides and the size of an angle that is not the included angle, we can still use the cosine rule to find the length of the other side. In this case we can either rearrange the formula or substitute the given values and solve an equation. a2 = – (2 × 7 × 4 × cos 48°) a2 = (to 2 d.p.) a = 5.25 cm (to 2 d.p.)

15 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 = b2 + c2 – a2 2bc cos A = – 82 2 × 4 × 6 Point out that if the cosine of an angle is negative, we expect the angle to be obtuse. This is because the cosine of angles in the second quadrant is negative. We do not have the same ambiguity as with the sine rule where the sine of angles in both the first and second quadrants are positive and so two solutions between 0° and 180° exist. Angles in a triangle can only be within this range. This is negative so A must be obtuse. cos A = –0.25 A = cos–1 –0.25 A = ° (to 2 d.p.)

16 Using the cosine rule to solve triangles
Be aware that the lengths of the sides and the angles have been rounded. Show the side lengths and angle sizes to a greater degree of accuracy to overcome this. Reveal the lengths of all three sides. Ask a volunteer to show how the cosine rule can be used to find the size of a required angle. Change the shape of the triangle by dragging on the vertices. Reveal the lengths of two of the sides and the angle between them. Ask a volunteer to show how the cosine rule can be used to find the length of the third side. Make the problem more difficult by revealing two sides and an angle other than the one between them.

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 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 B C 7 cm 4 cm 47° We can find the height h using the sine ratio. h h 4 = sin 47° 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
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 Talk through the formula as it is written in words. Remind students that the included angle is the angle between the two given sides. Remind students, too, that when labeling the sides and angles in a triangle it is common to label the vertices with capital A, B and C. The side opposite vertex A is labeled a, the side opposite vertex B is labeled b and the side opposite vertex C is labeled c. This formula could also be written as ½ bc sin A or ½ ac sin B. Area of triangle ABC = ab sin C 1 2

20 The area of a triangle using ½ ab sin C
Drag the vertices of the triangle to produce a variety of examples. The solution can be hidden or revealed. To vary the activity hide the angle or one of the sides by clicking on it. Reveal the area and ask students to find the missing angle or the length of the side. Be aware that the lengths of the sides and the angles have been rounded. This means that, for example, the three angles in the triangle may not add up to exactly 180°.

21 Contents 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
An angle is a measure of rotation. The system of using degrees to measure angles, where 1° is equal to 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. Another unit of rotation is the “gradian”, where 1 gradian is 1/400 of a full turn and so a right angle is 100 gradians. Most calculators have a ‘grad’ option although this unit is rarely used. For example, of a full turn is equal to 160°.

23 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. r 1 rad O 1 radian can be written as 1 rad or 1c. 2π rad = 360° So: rad =

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

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
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. r θ O A B 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: Point out that the size of the angle subtended by an arc is directly proportional to the length of the arc. Length of arc AB = θr where θ is measured in radians. When θ is measured in degrees the length of AB is

29 Finding the area of a sector
We can also find the area of a sector using radians. 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 2π radians. So the area of the sector AOB is of the area of the full circle.  Area of sector AOB = If necessary remind students that the area of a full circle is πr2. In general: Area of sector AOB = r2θ where θ is measured in radians. When θ is 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) length of arc AB = θr 9 cm O A B = 6π cm b) area of sector AOB = r2θ = 27π cm2

31 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. 5 cm 45° O A B The dots at the end of the number indicate that it has not been rounded but kept in the calculator. This is to avoid rounding errors. Occasionally the solution is required to be written exactly in which case it is left in terms of π. 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 = … – … = cm2 (to 3 sig. figs.) This formula does not have to be leant since it can be derived from the formulae for the area of a sector and the area of a triangle. Students sometimes confuse sectors and segments. Remember, a sector is like a slice of cake (both contain the letter c). In general, the area of a segment of a circle of radius r is: where θ is 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
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 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 = 2θ we 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 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 = if x = But x = 2θ, so: The second solution could also be found by considering the unit circle. θ = This is the complete solution set in the range

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
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: sin C 7 = 6 sin 50° The second solution is found by subtracting the first solution from 180°. sin B = 7 sin 50° 6 sin–1 B = 7 sin 50° 6 B = 63.34° or ° (to 2 d.p.)

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

39 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. O θ A B C D 6 cm 10 cm 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
a) Area of sector AOC = × 102 × θ = 50θ Area of sector BOD = × 62 × θ = 18θ Area of shaded region = 50θ – 18θ = 32θ b) 32θ = 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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