FOM & PreCalc 10 U7 Trigonometry

S3 Trigonometry Contents A S3.1 Right-angled triangles A S3.2 The three trigonometric ratios A S3.3 Finding side lengths A 7.4 Finding angles A S3.5 Angles of elevation and depression A S3.6 Trigonometry in 3-D

The inverse of sin 30° 0.5 sin θ = 0.5, what is the value of θ? To work this out use the sin–1 key on the calculator. sin–1 0.5 = 30° sin–1 is the inverse of sin. It is sometimes called arcsin. sin Make sure that pupils can locate the sin–1 key on their calculators. Stress that sin and sin–1 are inverse functions. sin 30° = 0.5 and sin–1 0.5 = 30°. Remind pupils of the use of –1 to denote the multiplicative inverse or reciprocal. This is an extension of this notation. sin–1 30° 0.5

The inverse of sin Make sure that pupils can locate the sin–1 key on their calculators. Stress that sin and sin–1 are inverse functions. sin 30° = 0.5 and sin–1 0.5 = 30°. Remind pupils of the use of –1 to denote the multiplicative inverse or reciprocal. This is an extension of this notation.

The inverse of cos 60° 0.5 Cos θ = 0.5, what is the value of θ? To work this out use the cos–1 key on the calculator. cos–1 0.5 = 60° Cos–1 is the inverse of cos. It is sometimes called arccos. cos Make sure that pupils can locate the cos–1 key on their calculators. Stress that cos and cos–1 are inverse functions. cos 60° = 0.5 and cos–1 0.5 = 60°. cos–1 60° 0.5

The inverse of tan 45° 1 tan θ = 1, what is the value of θ? To work this out use the tan–1 key on the calculator. tan–1 1 = 45° tan–1 is the inverse of tan. It is sometimes called arctan. tan Make sure that pupils can locate the tan–1 key on their calculators. Stress that tan and tan–1 are inverse functions. tan 45° = 1 and tan–1 1 = 45°. tan–1 45° 1

4 steps we need to follow: Step 1 Find which two sides we knowout of Opposite, Adjacent and Hypotenuse. Step 2 Use SOHCAHTOA to decide which one of Sine, Cosine or Tangent ratio to use in this question. Step 3 For Sine calculate Opposite/Hypotenuse, for Cosine calculate Adjacent/Hypotenuse or for Tangent calculate Opposite/Adjacent. Step 4 Find the angle from your calculator, using one of sin-1, cos-1 or tan-1 Make sure that pupils can locate the tan–1 key on their calculators. Stress that tan and tan–1 are inverse functions. tan 45° = 1 and tan–1 1 = 45°.

Finding angles Find θ to 2 decimal places. 5 cm 8 cm Find θ to 2 decimal places. We are given the lengths of the sides opposite and adjacent to the angle, so we use: tan θ = opposite adjacent On the calculator we can key in tan–1 (8 ÷ 5). This avoids rounding errors when the ratio cannot be written exactly as a decimal. tan θ = 8 5 θ = tan–1 (8 ÷ 5) = 57.99° (to 2 d.p.)

Finding angles Find θ to 2 decimal places. We are given the lengths of the sides opposite and hypotenuse to the angle, so we use: sin θ = opposite hypotenuse sin θ = 2.5 5 θ = sin–1 ( 2.5÷ 5) = 30.00° (to 2 d.p.)

Finding angles We are given the lengths of the sides opposite and adjacent to the angle, so we use: tan θ = 300 400 tan θ = opposite adjacent θ = tan–1 (300 ÷ 400) = 36.87° (to 2 d.p.)

Finding angles We are given the lengths of the sides hypotenuse and adjacent to the angle, so we use: cos θ = adjacent hypotenuse cos θ = 6750 8100 θ = cos–1 (6750 ÷ 8100) = 33.56° (to 2 d.p.)

Finding angles We are given the lengths of the sides opposite and hypotenuse to the angle, so we use: sin θ = opposite hypotenuse sin θ = 18.88 30 θ = sin–1 ( 18.88÷ 30) = 39.00° (to 2 d.p.)

Finding angles Use this activity to practice finding the size of angles given two sides in a right-angled triangle.

7.5 Angles of elevation and depression Contents Chapter 7 Trigonometry A 7.1 Right-angled triangles A 7.2 The three trigonometric ratios A 7.3 Finding side lengths A 7.4 Finding angles A 7.5 Angles of elevation and depression A 7.6 Trigonometry in 3-D

Angles of Elevation & Depression When we want to measure the height of an “inaccessible” object like a tree, pole, building, or cliff, we can utilize the concepts of trigonometry.

The Angle of Elevation The Angle of Elevation is the angle from the horizontal to your line of sight (ie. You are looking upwards at the object)

Example Problem Michael, whose eyes are six feet off the ground, is standing 36 feet away from the base of a building, and he looks up at a 50° angle to a point on the edge of building’s roof. To the nearest foot, how tall is the building? We are given the lengths of the adjacent side and the reference angle. We need to solve for the opposite side. tan θ = opposite adjacent tan 50°= OPP 360 OPP = 36 × tan 50° = 42.90 ft (to 2 d.p.)

Example Problem The sun is at an angle of elevation of 58°. A tree casts a shadow 20 meters long on the ground. How far is it from the top of the tree to the top of the shadow? We are given the lengths of the adjacent side and the reference angle. We need to solve for the hypotenuse. cos θ = adjacent hypotenuse cos 58°= 20 HYP HYP = 20 × cos 58° = 10.60m (to 2 d.p.)

Angles of elevation Explain that an angle of elevation is the angle made between the horizon and the top of an object that you are looking up at. Explain that we can use an instrument to measure angles of elevation. By measuring the angle of elevation from the ground to the top of a tall building we can apply trigonometry to determine the height of the building. Hide either the angle of elevation, the height of the building or the distance of the foot of the building to the point of measurement. Ask pupils to use the tangent ratio to find the unknown measurements. Change the position of the observer to change the measurements. Reset the activity to change the building.

The Angle of Depression The Angle of Depression is the angle from the horizontal to the line of sight. (i.e. you are looking downwards at the object)

Example Problem From the top of a vertical cliff 40 m high, the angle of depression of an object that is level with the base of the cliff is 34º. How far is the object from the base of the cliff? We are given the lengths of the adjacent side and the reference angle. We need to solve for the opposite side, x. tan θ = opposite adjacent tan 90°-34°= OPP 40 OPP = 40 × tan 56° = 59.30m (to 2 d.p.)

Example Problem We are given the lengths of the adjacent side and the reference angle. We need to solve for the hypotenuse. HYP = 20 × cos 58° cos θ = adjacent hypotenuse cos 58°= 20 HYP = 10.60m (to 2 d.p.)

Angles of depression Explain that an angle of depression is the angle made between the horizon and the bottom of an object that you are looking down on. By measuring the angle of depression from the the top of a cliff, for example, we can apply trigonometry to determine the distance of objects from the foot of the cliff. Hide either the angle of depression, the distance between the boat and the viewer or the distance between the boat and the foot of the cliff. Ask pupils to use the tangent ratio to find the unknown measurements. Modify the values by changing the position of the boat.

S3 Trigonometry Contents A S3.1 Right-angled triangles A S3.2 The three trigonometric ratios A S3.3 Finding side lengths A S3.4 Finding angles A S3.5 Angles of elevation and depression A S3.6 Trigonometry in 3-D

Angles in a cuboid Stress that when solving problems involving right-angled triangles in 3-D, we should always draw a 2-D diagram of the right-angled triangle in question, labelling all the known sides and angles. Show how angle AGB can be found by first using Pythagoras’ theorem to find the length of BG. Once we know the length of BG we can find angle AGB using the tan ratio. Reveal the answer. If there is a small discrepancy, this is due to rounding. Change the size of the cuboid and ask a volunteer to calculate the new size of angle AGB.

Lengths in a square-based pyramid Tell pupils that the point M is directly below point E. Point E can be moved to change the angle between the base and the sloping side. Demonstrate how to find the length of EM using the sine ratio. Reveal the answer. If there is a small discrepancy, this is due to rounding. Change the height of the pyramid by dragging on point E and ask a volunteer to calculate the new pyramid height EM. Ask pupils if it is possible to find the length of the side of the square base AB. Challenge pupils to find the length of AB using a combination of trigonometry and Pythagoras’ theorem.