The graph of sin θ Teacher notes Trace out the shape of the sine curve and note its properties. The curve repeats itself every 360°. Also, –1 < sin θ < 1. Important to come up with as many patterns as possible for this, particularly: sinx = –sin(–x) sinx = sin(x+360) sinx = sin(180–x)

The graph of cos θ Teacher notes Trace out the shape of the cosine curve and note its properties. The curve repeats itself every 360°. Also, –1 < sin θ < 1. The cosine curve is symmetrical about the vertical axis. Ask pupils to come up with as many patterns as possible for this, particularly: cosx = cos(–x) cosx = cos(x+360) sinx = sin(360 –x) Students should also be guided to look at the similarities with the sine curve and the connection: sinx = cos(x-90) or cosx = sin(x+90)

The graph of tan θ Teacher notes Trace out the shape of the tangent curve and note its properties. The curve repeats itself every 180°. Tan θ is undefined at 90° (and 180n + 90° for integer values of n). It is important to come up with as many patterns as possible for this, particularly: tanx = –tan(–x) tanx = tan(x+180)

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

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

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. What is the area of triangle ABC? h = the height of the triangle. A 4 cm h is found using the sine ratio. h h 47° = sin 47° B C 4 Teacher notes Remind pupils that the included angle is the angle between two given sides. 7 cm h = 4 sin 47° 1 Area of triangle ABC = × base × height 2 1 = × 7 × 4 sin 47° 2 = 10.2 cm2 (to 1 d.p.)

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 Teacher notes 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 labelled a, the side opposite vertex B is labelled b and the side opposite vertex C is labelled c. This formula could also be written as ½ bc sin A or ½ ac sin B. a 1 Area of triangle ABC = ab sin C 2

The area of a triangle using ½ ab sin c Teacher notes 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 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°.

Using the area formula A triangle has an area of 10m2. It has 2 lengths of 8 m and 5 m. What angles could be between the sides? 22° 54° 130° 2.8 m 121 cm 14 cm 72 cm 2.4 m 18 cm Teacher notes A triangle with area of 10m2 and 2 lengths of 8m and 5m would have an angle of: 30° or 150° Triangle 1: 2.72m2 Triangle 2: The right angled triangle needs the base to be calculated (97.25) before finding the area of 3501cm2 Triangle 3:The last one needs the included angle (28) to find the area of 59.2cm2 Find the areas of the 3 triangles.

The sine rule Any triangle (ABC) can be divided two right-angled triangles, ACD and BDC, if a perpendicular line, h, is drawn from AB to C. C a is the side opposite A and b is the side opposite B. b a h A B D Teacher notes We call the perpendicular h for height (not h for hypotenuse). h h sin A = sin B = b a h = b sin A h = a sin B b sin A = a sin B

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 a = sin B 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 c = sin B sin C

The sine rule C b a A B c a b c sin A sin B sin C = = or = = sin A Teacher notes We can use the first form of the formula to find side lengths and the second form of the equation to find angles. It should be emphasised that the first form of the formula is given on the formula sheet in the answer book of the exam. a b c sin A sin B sin C = = or = = sin A sin B sin C a b c

Using the sine rule to find side lengths If given two angles in a triangle and the length of a side opposite one of the angles, use the sine rule to find the length of the side opposite the other angle. Find the length of side a. a 7 cm 118° 39° A B C Using the sine rule, a 7 = sin 118° sin 39° Teacher notes 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 = 9.82 cm (to 2 d.p.)

Using the sine rule to find side lengths Teacher notes 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.

Using the sine rule to find angles If given two side lengths in a triangle and the angle opposite one of the given sides, use the sine rule to find the angle opposite the other given side. Find the angle at B. 6 cm 46° B 8 cm A C Using the sine rule, sin B 8 sin 46° = 6 Teacher notes When trying to find an angle it is easier to use the formula in the form sin A/a = sin B/b. Tell students that many of them may find it easier to write down the answer to sinB after calculating, before finding sin–1. The second possible solution will be found on the following slide. sin B = 8 sin 46° 6 sin–1 B = 8 sin 46° 6 B = 73.56° (to 2 d.p.)

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 sin θ = sin (180° – θ) 46° 6 cm B For every acute solution, there is a corresponding obtuse solution. Teacher notes Remind pupils 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 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° = 106.44° (to 2 d.p.)

Using the sine rule to find angles Teacher notes 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.

The cosine rule Consider any triangle ABC. If we drop a perpendicular line, 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 Teacher notes 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.

The cosine rule Using Pythagoras’ theorem in triangle ACD, C b2 = x2 + h2 1 b a h Also, x cos A = A B b x D c – x x = b cos A 2 In triangle BCD, a2 = (c – x)2 + h2 Teacher notes 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 a2 = c2 – 2cb cos A + b2 a2 = b2 + c2 – 2bc cos A

The cosine rule For any triangle ABC, A B C c a b a2 = b2 + c2 – 2bc cos A or Teacher notes We can use the first form of the formula to find side lengths and the second form of the equation to find angles. Remind students that the first form of the equation is given in the formula sheet, but that the cosA version is not. They will need to be able to rearrange the first version into the second version. cos A = b2 + c2 – a2 2bc

Using the cosine rule to find side lengths If given the length of two sides in a triangle and the size of the angle between them, the cosine rule can be used to find the length of the other side. Find the length of side a. B C A 7 cm 4 cm 48° a a2 = b2 + c2 – 2bc cos A a2 = 72 + 42 – 2 × 7 × 4 × cos 48° Teacher notes 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 or substitute the given values and solve an equation. a2 = 27.53 (to 2 d.p.) a = 5.25 cm (to 2 d.p.)

Using the cosine rule to find angles If given the lengths of all three sides in a triangle, the cosine rule can be used 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 = 42 + 62 – 82 2 × 4 × 6 Teacher notes 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 = 104.48° (to 2 d.p.)

Using the cosine rule to find angles Teacher notes 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.

Using the rules Using the rules An aeroplane starts at the airport, A, and flies 8.5 km on a bearing of 060°. 160° Using the rules 8.5 km It then gets blown off course on a bearing of 160° for 7 km. 060° 7 km A Teacher notes Find the angle between the 8.5km and 7km parts and use the cosine rule. Use the sine rule for the angle between the 7km and home direction. Angle at top of triangle is 80°. Distance home is 10.03km. Angle on bottom right of triangle is 56.6°. Bearing is 283.4. Photo credits: © Luis Louro 2010, Shutterstock.com The captain decides to return home. How far does she have to fly and in what direction?

Using the rules A surveyor is trying to calculate the height of a tower. He measures the angle of elevation at 32°. He then moves closer by 15m, and the angle is now 42°. 42° 32° Teacher notes Find all the angles in the right triangle and use the sine rule. Use SOHCAHTOA in the left triangle. Angle at top of left triangle is 48°. Length of hypotenuse shared with the left triangle is 45.8m. Height of tower is 30.6m. 15 m Find the height of the tower.