FOM Pre Calc 10 U 7 Trigonometry 1
- Slides: 27
FOM & Pre. Calc 10 U 7 Trigonometry 1 of 43 © Boardworks Ltd 2008
Contents S 3 Trigonometry A S 3. 1 Right-angled triangles A S 3. 2 The three trigonometric ratios A S 3. 3 Finding side lengths A 7. 4 Finding angles A S 3. 5 Angles of elevation and depression A S 3. 6 Trigonometry in 3 -D 2 of 43 © Boardworks Ltd 2008
The inverse of 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 30° 0. 5 sin– 1 3 of 43 © Boardworks Ltd 2008
The inverse of sin 4 of 43 © Boardworks Ltd 2008
The inverse of cos 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 60° 0. 5 cos– 1 5 of 43 © Boardworks Ltd 2008
The inverse of 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 45° 1 tan– 1 6 of 43 © Boardworks Ltd 2008
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 7 of 43 © Boardworks Ltd 2008
Finding angles 8 cm Find θ to 2 decimal places. 5 cm θ We are given the lengths of the sides opposite and adjacent to the angle, so we use: opposite tan θ = adjacent 8 tan θ = 5 θ = tan– 1 (8 ÷ 5) = 57. 99° (to 2 d. p. ) 8 of 43 © Boardworks Ltd 2008
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: opposite sin θ = hypotenuse 2. 5 sin θ = 5 θ = sin– 1 ( 2. 5÷ 5) = 30. 00° (to 2 d. p. ) 9 of 43 © Boardworks Ltd 2008
Finding angles We are given the lengths of the sides opposite and adjacent to the angle, so we use: opposite tan θ = adjacent 300 tan θ = 400 θ = tan– 1 (300 ÷ 400) = 36. 87° (to 2 d. p. ) 10 of 43 © Boardworks Ltd 2008
Finding angles We are given the lengths of the sides hypotenuse and adjacent to the angle, so we use: adjacent cos θ = hypotenuse 6750 cos θ = 8100 θ = cos– 1 (6750 ÷ 8100) = 33. 56° (to 2 d. p. ) 11 of 43 © Boardworks Ltd 2008
Finding angles We are given the lengths of the sides opposite and hypotenuse to the angle, so we use: opposite sin θ = hypotenuse sin θ = 18. 88 30 θ = sin– 1 ( 18. 88÷ 30) = 39. 00° (to 2 d. p. ) 12 of 43 © Boardworks Ltd 2008
Finding angles 13 of 43 © Boardworks Ltd 2008
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 14 of 43 © Boardworks Ltd 2008
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. 15 of 43 © Boardworks Ltd 2008
The Angle of Elevation is the angle from the horizontal to your line of sight (ie. You are looking upwards at the object) 16 of 43 © Boardworks Ltd 2008
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. opposite tan θ = adjacent tan 50°= OPP 360 OPP = 36 × tan 50° = 42. 90 ft (to 2 d. p. ) 17 of 43 © Boardworks Ltd 2008
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. 20 adjacent cos 58°= cos θ = hypotenuse HYP = 20 × cos 58° = 10. 60 m (to 2 d. p. ) 18 of 43 © Boardworks Ltd 2008
Angles of elevation 19 of 43 © Boardworks Ltd 2008
The Angle of Depression is the angle from the horizontal to the line of sight. (i. e. you are looking downwards at the object) 20 of 43 © Boardworks Ltd 2008
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. opposite tan θ = adjacent tan 90°-34°= OPP = 40 × tan 56° 40 = 59. 30 m (to 2 d. p. ) 21 of 43 © Boardworks Ltd 2008
Example Problem We are given the lengths of the adjacent side and the reference angle. We need to solve for the hypotenuse. 20 adjacent cos 58°= cos θ = hypotenuse HYP 22 of 43 HYP = 20 × cos 58° = 10. 60 m (to 2 d. p. ) © Boardworks Ltd 2008
Angles of depression 23 of 43 © Boardworks Ltd 2008
24 of 43 © Boardworks Ltd 2008
Contents S 3 Trigonometry A S 3. 1 Right-angled triangles A S 3. 2 The three trigonometric ratios A S 3. 3 Finding side lengths A S 3. 4 Finding angles A S 3. 5 Angles of elevation and depression A S 3. 6 Trigonometry in 3 -D 25 of 43 © Boardworks Ltd 2008
Angles in a cuboid 26 of 43 © Boardworks Ltd 2008
Lengths in a square-based pyramid 27 of 43 © Boardworks Ltd 2008
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