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**S4 Further trigonometry**

KS4 Mathematics S4 Further trigonometry

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**S4.1 Sin, cos and tan of any angle**

Contents S4 Further trigonometry A S4.1 Sin, cos and tan of any angle A S4.2 Sin, cos and tan of 30°, 45° and 60° A S4.3 Graphs of trigonometric functions A S3.4 Area of a triangle using ½ab sin C A S3.5 The sine rule A S4.6 The cosine rule

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**The opposite and adjacent sides**

Suppose we have a right-angled triangle with hypotenuse h and acute angle θ. θ h a) Write an expression for the length of the opposite side in terms of h and θ. b) Write an expression for the length of the adjacent side in terms of h and θ. The following slides will show that the length of the opposite side in a right-angled triangle can be written as h sin θ and that the adjacent side in a right-angle triangle can be written as h cos θ . From this it follows that when the hypotenuse is of unit length, the opposite side in a right-angled triangle can be written as sin θ and that the adjacent side can be written as cos θ . The values of sin θ and cos θ are then examined for right-angled triangles drawn on a coordinate grid.

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**The opposite and adjacent sides**

Suppose we have a right-angled triangle with hypotenuse h and acute angle θ. θ h a) sin θ = opp hyp b) cos θ = adj hyp opp = hyp × sin θ adj = hyp × cos θ opp = h sin θ adj = h cos θ

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**The opposite and adjacent sides**

So, for any right-angled triangle with hypotenuse h and acute angle θ. We can label the opposite and adjacent sides as follows: h h sin θ θ h cos θ Establish that tan θ = sin θ / cos θ for all values of θ . opposite tan θ = h sin θ h cos θ We can write, adjacent tan θ = sin θ cos θ

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The sine of any angle Explain that we can find the sine of any angle by considering the movement of the point P which is fixed at 1 unit from the origin on a coordinate grid. Sin θ is given by the y-coordinate of the point P. This is show by the length of the bold orange line. Explain that moving the point P in an anticlockwise direction increases the angle between OP and the x-axis. In maths, an anti-clockwise rotation is a positive rotation. Slowly, drag the point P through 360°, starting at 0° and ending at 360°. Observe the length of the line representing the y-coordinate of the point P as it is rotated. Point out that when the line is in the first and the second quadrant, that is, when θ is between 0° and 180°, it is above the x-axis and therefore positive. In other words, the sin of angles between 0° and 180°, is positive. When the line representing sin θ is in the third and the fourth quadrant, that is, when θ is between 180° and 360°, it is below the x-axis and therefore negative. In other words, the sine of angles between 180° and 360° is negative. In this activity, P can be moved through any angle between –360° and 720°. Demonstrate the sine of these angles if required. Explain that P can be moved through any positive or negative angle in this way. Focusing on angles between 0° and 180° (that is, angles in the first and second quadrants) demonstrate the relationships between pairs of angles that have the same sine. For example, show that sin 32° = sin 148°. Conclude that sin θ = sin (180° – θ). Focus next on angles between 180° and 270° (that is, angles in the third quadrant) . By looking at the sine of the associated acute angle show that sin θ = – sin (θ – 180°). In the fourth quadrant, examine angles between 270° and 360° and between 0° and –90°. Conclude that sin θ = – sin (360° – θ) for angles between 270° and 360° and sin –θ = – sin θ.

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**Sine of angles in the second quadrant**

We have seen that the sine of angles in the first and second quadrants are positive. The sine of angles in the third and fourth quadrants are negative. In the second quadrant, 90° < θ < 180°. sin θ = sin (180° – θ) Stress that the sine of any obtuse angle θ is equal to the sine of its supplementary angle (180° – θ). In the second quadrant θ can also be between –180° and –270°. In general, in the second quadrant ( n)° < θ < ( n)° for any integer n. Ask pupils to verify, using their calculators, that sin 130° = sin 50°. For example, sin 130° = sin (180° – 130°) = sin 50° = (to 3 sig. figs)

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**Sine of angles in the third quadrant**

sin θ = –sin (θ – 180°) For example, sin 220° = – sin (220° – 180°) = – sin 40° In the third quadrant θ can also be between –90° and –180°. In general, in the third quadrant ( n)° < θ < ( n)° for any integer n. = – (to 3 sig. figs) Verify, using a scientific calculator, that sin 220° = –sin 40°

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**Sine of angles in the fourth quadrant**

In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90° sin θ = –sin(360° – θ) or sin –θ = –sin θ For example, sin 300° = –sin (360° – 300°) = –sin 60° = –0.866 (to 3 sig. figs) In general, in the fourth quadrant ( n)° < θ < 360n° for any integer n. Ask pupils to verify these answers using a calculator. sin –35° = –sin 35° = –0.574 (to 3 sig. figs)

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The cosine of any angle Explain that cos θ is given by the x-coordinate of the point P. This is shown by the length of the bold orange line. Slowly, drag the point P through 360°, starting at 0° and ending at 360°. Observe the length of the line representing the x-coordinate of the point P as it is rotated. Point out that when the line is in the first and the fourth quadrants, that is, when θ is between 0° and 90° or between 270° and 360° (also between 0° and –90°), it is to the right of the y-axis and therefore positive. When the line representing cos θ is in the second and third quadrants, that is, when θ is between 90° and 270°, it is to the left the y-axis and therefore negative. In other words, the cosine of angles between 90° and 270°, is negative. In this activity, P can be moved through any angle between –360° and 720°. Demonstrate the cosine of these angles if required. Explain that P can be moved through any positive or negative angle in this way. Focusing on angles between 0° and 180° (that is, angles in the first and second quadrants) demonstrate the relationships between pairs of angles that have the same cosine, but are of opposite sign. For example, show that cos 148° = –cos 32°. Conclude that cos θ = –cos (180° – θ). Focus next on angles between 180° and 270° (that is, angles in the third quadrant) . Show that cos θ = – cos (θ – 180°). In the fourth quadrant, examine angles between 270° and 360° and between 0° and –90°. Conclude that cos θ = cos (360° – θ) for angles between 270° and 360° and cos θ = cos –θ for angles between 0° and –90°.

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**Cosine of angles in the second quadrant**

We have seen that the cosines of angles in the first and fourth quadrants are positive. The cosines of angles in the second and third quadrants are negative. In the second quadrant, 90° < θ < 180°. cos θ = –cos (180° – θ) Stress that the cosine of any obtuse angle θ is equal to the negative cosine of its supplementary angle (180° – θ). In the second quadrant θ can also be between –180° and –270°. In general, in the second quadrant ( n)° < θ < ( n)° for any integer n. Ask pupils to verify, using their calculators, that cos 100° = –cos 80°. For example, cos 100° = –cos (180° – 100°) = –cos 80° = –0.174 (to 3 sig. figs)

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**Cosine of angles in the third quadrant**

cos θ = –cos (θ – 180°) For example, cos 250° = –cos (250° – 180°) = –cos 70° In the third quadrant θ can also be between –90° and –180°. In general, in the third quadrant ( n)° < θ < ( n)° for any integer n. = –0.342 (to 3 sig. figs.) Verify, using a scientific calculator, that cos 250° = –cos 70°

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**Sine of angles in the fourth quadrant**

In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90° cos θ = cos(360° – θ) or cos –θ = cos θ For example, cos 317° = cos (360° – 317°) = cos 43° = (to 3 sig. figs.) Remind pupils that the cosine of any angle the fourth quadrant is always positive. In general, in the fourth quadrant ( n)° < θ < 360n° for any integer n. Ask pupils to verify these answers using a calculator. cos –28° = cos 28° = (to 3 sig. figs.)

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**The tangent of any angle**

Remind pupils that tan θ is given by sin θ/ cos θ. Tan θ is therefore given by the y-coordinate of the point P divided by the x-coordinate of the point P. Slowly, drag the point P through 360°, starting at 0° and ending at 360°. Observe the change in the value of tan θ as the point P is rotated. Point out that when P is in the first and the third quadrants, that is, when θ is between 0° and 90° or between 180° and 270°, the sin and cosine of the required angle is of the same sign. The tangent is therefore positive in these quadrants. The tangent of 90° and 270° is undefined when cos θ = 0 (because we cannot divide by 0). When the point P is in the second and fourth quadrants, that is, when θ is between 90° and 180° and between 270° and 360° , the sin and cosine of the required angle is of different sign. The tangent is therefore negative in these quadrants. Focusing on angles between 0° and 180° (that is, angles in the first and second quadrants) demonstrate the relationships between pairs of angles that have the same tangent, but are of opposite sign. For example, show that tan 127° = –tan 53°. Conclude that tan θ = –tan (180° – θ). Focus next on angles between 180° and 270° (that is, angles in the third quadrant) . Show that tan θ = tan (θ – 180°). In the fourth quadrant, examine angles between 270° and 360° and between 0° and –90°. Conclude that tan θ = –tan (360° – θ) for angles between 270° and 360° and tan –θ = –tan θ for angles between 0° and –90°.

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**The tangent of any angle**

This demonstration shows more clearly how the value of tan θ varies as θ varies. In particular, it can be shown that tan 90° and tan 270° are undefined. This is because the tangent is parallel to the x-axis at these points. The circle has radius 1. Explain to pupils that the length of the tangent at P from P to the x-axis gives us the value of tan θ. Remind pupils that the tangent of a circle forms a right angle with the radius. We therefore have a right-angled triangle with opposite side of length tan θ and adjacent side of length 1. Opposite/adjacent = tan θ as required. Slowly move the point P through 0° < θ < 360° and observe how the value of tan θ varies. Draw pupils attention to the fact that when θ is close to 90° and 270° it get very large very quickly. Establish that at these points the tangent at P is parallel to the x-axis. Since the tangent will never meet the x-axis at 90° or 270° tan θ is undefined at these angles. The slope of the tangent defines whether it is positive or negative. A positive gradient gives a negative value and a negative gradient gives a positive value.

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**Tangent of angles in the second quadrant**

We have seen that the tangent of angles in the first and third quadrants are positive. The tangent of angles in the second and fourth quadrants are negative. In the second quadrant, 90° < θ < 180°. tan θ = –tan (180° – θ) Stress that the tangent of any obtuse angle θ is equal to the negative tangent of its supplementary angle (180° – θ). In the second quadrant θ can also be between –180° and –270°. In general, in the second quadrant ( n)° < θ < ( n)° for any integer n. Ask pupils to verify, using their calculators, that cos 116° = –tan 2.05°. For example, tan 116° = –tan (180° – 116°) = –tan 64° = –2.05 (to 3 sig. figs)

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**Tangent of angles in the third quadrant**

tan θ = tan (θ – 180°) For example, tan 236° = tan (236° – 180°) = tan 56° Stress that the tangent of any angle in the third quadrant is positive. In the third quadrant θ can also be between –90° and –180°. In general, in the third quadrant ( n)° < θ < ( n)° for any integer n. = 1.48 (to 3 sig. figs) Verify, using a scientific calculator, that tan 236° = tan 56°

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**Tangent of angles in the fourth quadrant**

In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90° tan θ = –tan(360° – θ) or tan –θ = –tan θ For example, tan 278° = –tan (360° – 278°) = –tan 82° = –7.12 (to 3 sig. figs) In general, in the fourth quadrant ( n)° < θ < 360n° for any integer n. Ask pupils to verify these answers using a calculator. tan –16° = –tan 16° = –0.287 (to 3 sig. figs)

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**Sin, cos and tan of angles between 0° and 360°**

The sin, cos and tan of angles in the first quadrant are positive. In the second quadrant: sin θ = sin (180° – θ) cos θ = –cos (180° – θ) tan θ = –tan (180° – θ) In the third quadrant: sin θ = –sin (θ – 180°) cos θ = –cos (θ – 180°) tan θ = tan (θ – 180°) This slide summarizes the results established for the sin, cos and tan of angles between 0° and 360°. In the fourth quadrant: sin θ = –sin (360° – θ) cos θ = cos (360° – θ) tan θ = –tan(180° – θ)

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Remember CAST We can use CAST to remember in which quadrant each of the three ratios are positive. 2nd quadrant 1st quadrant Sine is positive S All are positive A 3rd quadrant 4th quadrant Introduce pupils to CAST to help them remember in which quadrant each of the thee ratios are positive. Tangent is positive T Cosine is positive C

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Positive or negative? Start by establishing which quadrant the given angle is in. Ask pupils to use this to decide whether the sin, cos or tan of the given angle will be positive or negative.

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**Find the equivalent ratio**

For each example find three equivalent ratios.

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Solving equations in θ Pupils can use their calculators to find a value for θ between 0° and 90°. Pupils should then be encouraged to recall in which quadrant the given ratio is positive (or negative). They can then use a sketch of the four quadrants to find the other three solutions in the given range. These equations can also be solved using sine, cosine or tangent graphs.

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**S4.2 Sin, cos and tan of 30°, 45° and 60°**

Contents S4 Further trigonometry A S4.1 Sin, cos and tan of any angle A S4.2 Sin, cos and tan of 30°, 45° and 60° A S4.3 Graphs of trigonometric functions A S3.4 Area of a triangle using ½ab sin C A S3.5 The sine rule A S4.6 The cosine rule

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Sin, cos and tan of 45° A right-angled isosceles triangle has two acute angels of 45°. Suppose the equal sides are of unit length. 45° 2 1 Using Pythagoras’ theorem, The hypotenuse = 1² + 1² 45° = 2 1 Make sure that pupils can see that these ratios would be the same for any right-angled isosceles triangle. If the equal sides were of a different length, for example 3, the hypotenuse would be of length 32. In each ratio the 3’s would cancel and so simplify to those shown above. If required, verify using a scientific calculator that sin and cos of 45° = … = 1/2 Stress that 1/2 is an exact answer. Sin and cos of 45° cannot be written exactly as a decimal. We can use this triangle to write exact values for sin, cos and tan 45°: sin 45° = 1 2 cos 45° = 1 2 tan 45° = 1

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Sin, cos and tan of 30° Suppose we have an equilateral triangle of side length 2. 2 60° 30° 1 2 60° If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. 3 Using Pythagoras’ theorem, The height of the triangle = 2² – 1² = 3 We can use this triangle to write exact values for sin, cos and tan 30°: sin 30° = 1 2 cos 30° = 3 2 tan 30° = 1 3

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Sin, cos and tan of 60° Suppose we have an equilateral triangle of side length 2. 2 60° 30° 1 If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. 3 Using Pythagoras’ theorem, The height of the triangle = 2² – 1² = 3 We can also use this triangle to write exact values for sin, cos and tan 60°: sin 60° = 3 2 cos 60° = 1 2 tan 60° = 3

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**Sin, cos and tan of 30°, 45° and 60°**

The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: 30° sin cos tan 45° 60° 1 2 1 2 3 2 3 2 1 2 1 2 1 3 1 3 Ask pupils to give the exact values of cos 150° and tan 150°. Ask pupils for sin, cos and tan of other angles associated with 30° (for example, 210°, 330°, 390°, – 30° or –150°). Use CAST to remind pupils in which quadrant each ratio is positive. See slide 20. Use this table to write the exact value of sin 150°: 1 2 sin 150° =

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**Sin, cos and tan of 30°, 45° and 60°**

The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: 30° sin cos tan 45° 60° 1 2 1 2 3 2 3 2 1 2 1 2 1 3 1 3 Ask pupils to give the exact values of sin 135° and tan 135°. Ask pupils for sin, cos and tan of other angles associated with 45° (for example, 225°, 315°, 405°, – 45° or –135°). Use CAST to remind pupils in which quadrant each ratio is positive. See slide 20. Use this table to write the exact value of cos 135°: –1 2 cos 135° =

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**Sin, cos and tan of 30°, 45° and 60°**

The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: 30° sin cos tan 45° 60° 1 2 1 2 3 2 3 2 1 2 1 2 1 3 1 3 Ask pupils to give the exact values of sin 120° and cos 120°. Ask pupils for sin, cos and tan of other angles associated with 60° (for example, 240°, 300°, 420°, – 60° or –120°). Use CAST to remind pupils in which quadrant each ratio is positive. See slide 20. Use this table to write the exact value of tan 120°: tan 120° = –3

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**Sin, cos and tan of 30°, 45° and 60°**

Write the following ratios exactly: 1 2 1) cos 300° = 2) tan 315° = –1 1 2 3) tan 240° = 3 4) sin –330° = 3 2 5) cos –30° = 6) tan –135° = 1 Ask pupils to complete this exercise individually before revealing the answers. Suggest to pupils that they write down which quadrant each angle falls into before writing down their solutions. –1 2 1 2 7) sin 210° = 8) cos 315° =

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**S4.3 Graphs of trigonometric functions**

Contents S4 Further trigonometry A S4.1 Sin, cos and tan of any angle A S4.2 Sin, cos and tan of 30°, 45° and 60° A S4.3 Graphs of trigonometric functions A S3.4 Area of a triangle using ½ab sin C A S3.5 The sine rule A S4.6 The cosine rule

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The graph of sin θ Trace out the shape of the sine curve and note its properties. The curve repeats itself every 360°. Also, –1 < sin θ < 1.

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The graph of cos θ Trace out the shape of the cosine curve and note it properties. The curve repeats itself every 360°. Also, –1 < sin θ < 1. The cosine curve is symmetrical about the vertical axis.

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The graph of tan θ Trace out the shape of the tangent curve and note it properties. The curve repeats itself every 180°. Tan θ is undefined at 90° (and 180n + 90° for integer values of n).

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**Transforming trigonometric graphs**

Use this activity to explore transformations of sine, cosine and tangent graphs. Transformations of functions is covered in more detail in A9 Graphs of non-linear functions. The input is x here rather than θ to be consistent with general function notation.

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**S3.4 Area of a triangle using ½ab sin C**

Contents S4 Further trigonometry A S4.1 Sin, cos and tan of any angle A S4.2 Sin, cos and tan of 30°, 45° and 60° A S4.3 Graphs of trigonometric functions A S3.4 Area of a triangle using ½ab sin C A S3.5 The sine rule A S4.6 The cosine rule

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**The area of a triangle Remember, h b 1 Area of a triangle = bh 2**

Remind pupils that the area of a triangle can be found by halving the length of the base multiplied by the perpendicular height. Area of a triangle = bh 1 2

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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? Let’s call the height of the triangle h. A B C 7 cm 4 cm 47° We can find 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.)

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**The area of a triangle using ½ ab sin C**

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 pupils that the included angle is the angle between the two given sides. Remind pupils, 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

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**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 of revealed. To vary the activity hide the angle or one of the sides by clicking on it. Reveal the area and ask pupils to find the missing angle or side length. 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 180°.

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**S4 Further trigonometry**

Contents S4 Further trigonometry A S4.1 Sin, cos and tan of any angle A S4.2 Sin, cos and tan of 30°, 45° and 60° A S4.3 Graphs of trigonometric functions A S3.4 Area of a triangle using ½ab sin C A S3.5 The sine rule A S4.6 The cosine rule

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**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

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**The sine rule b sin A = a sin B**

Dividing both sides of the equation by sin A and then by sin B we have: 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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**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 formula 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

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**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 formula in the form a/sin A = b/sin B. Encourage pupils 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 = (to 2 d.p.)

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**Using the sine rule to find side lengths**

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°. Reveal an angle and the side opposite it. Reveal one more angle and ask pupils to find the side opposite it by using the sine rule. Generate a new example by modifying the shape of the triangle.

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**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 formula 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.)

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**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 pupils that the sine of angles in the second quadrant, that is angle 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 pupils 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. B = 73.56° (to 2 d.p.) or B = 180° – 73.56° = ° (to 2 d.p.)

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**Using the sine rule to find angles**

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°. Reveal an angle and the side opposite it. Reveal one more side and ask pupils to find the angle opposite it by using the sine rule. Generate a new example by modifying the shape of the triangle.

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**S4 Further trigonometry**

Contents S4 Further trigonometry A S4.1 Sin, cos and tan of any angle A S4.2 Sin, cos and tan of 30°, 45° and 60° A S4.3 Graphs of trigonometric functions A S3.4 Area of a triangle using ½ab sin C A S3.5 The sine rule A S4.6 The cosine rule

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**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 we call the length AD x, then the length BD can be written as c – x.

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**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 , 1 2 This is the cosine rule. a2 = c2 – 2cb cos A + b2 a2 = b2 + c2 – 2bc cos A

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**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 formula to find side lengths and the second form of the equation to find angles.

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**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 pupils not to forget to find the square root. The answer should look sensible considering the other lengths. Advise pupils 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 of 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.)

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**Using the cosine rule to find side lengths**

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°. 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.

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**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.)

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**Using the cosine rule to find angles**

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°. Change the shape of the triangle by dragging on the vertices. 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.

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∆ABC has three angles… › ∡C is a right angle › ∡A and ∡B are acute angles We can make ratios related to the acute angles in ∆ABC A CB 3 4 5.

∆ABC has three angles… › ∡C is a right angle › ∡A and ∡B are acute angles We can make ratios related to the acute angles in ∆ABC A CB 3 4 5.

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