Chapter 2 Trigonometry. § 2.1 The Tangent Ratio TOA x Hypotenuse (h) Opposite (o) Adjacent (a) x Hypotenuse (h) Opposite (o) Adjacent (a) Hypotenuse.

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Presentation transcript:

Chapter 2 Trigonometry

§ 2.1 The Tangent Ratio

TOA x Hypotenuse (h) Opposite (o) Adjacent (a) x Hypotenuse (h) Opposite (o) Adjacent (a) Hypotenuse Opposite Adjacent

Example #1 Determine the tangent ratio for the following. Hypotenuse (h) Opposite (o) Adjacent (a)

Example #2 Determine the measures of to the nearest tenth of a degree. Hypotenuse (h) Opposite (o) Adjacent (a)

Example #3 The latitude of Fort Smith, NWT, is approximately 60 o. Determine whether this design for a solar panel is the best for Fort Smith. Justify your answer. h a o x = ° The best angle of inclination for the solar panel is the same as the latitude (60°) x This is not the best design for the solar panel because it is not equal to the latitude of Fort Smith (60 °)

Example #4 A 10-ft. ladder leans against the side of a building that is 7 ft. tall. What angle, to the nearest degree, does the ladder make with the ground? h a O 10 7 x a 2 + b 2 = c 2 a 2 + o 2 = h 2 a 2 = 51 a = 100 a = √ 51 a = 10 2 √ 51 on your calculator: x = tan -1 (7÷ √ (51)) <x = <x = 44°

§ 2.2 Using the Tangent Ratio to Calculate Lengths

The Tangent Ratio TOA x Hypotenuse (h) Opposite (o) Adjacent (a) x Hypotenuse (h) Opposite (o) Adjacent (a) Hypotenuse Opposite Adjacent

Example #1 Determine the length of AB to the nearest tenth of a centimeter Hypotenuse (h) Opposite (o) Adjacent (a)

Example #2 (first way) Determine the length of EF to the nearest tenth of a centimeter. h O a

Example #2 (second way) Determine the length of EF to the nearest tenth of a centimeter. h O a Angles in a triangle add up to 90° 70°

Example #3 A searchlight beam shines vertically on a cloud. At a horizontal distance of 250m from the searchlight, the angle between the ground and the line of sight to the cloud is 75 o. Determine the height of the cloud to the nearest metre. 250 m h O a The cloud is 933 metres above the ground

§ 2.4 The Sine and Cosine Ratios

The Sine & Cosine Ratios x Hypotenuse (h) Opposite (o) Adjacent (a) x Hypotenuse (h) Opposite (o) Adjacent (a) Hypotenuse Opposite Adjacent SOH CAH

Example #1 a) In identify the side opposite, the side adjacent to and the hypotenuse. b) Determine using sin and cos to the nearest hundredth Hypotenuse Opposite Adjacent

Example #2 Determine the measures of and to the nearest tenth of a degree. Hypotenuse Opposite Adjacent

Example #3 A water bomber is flying at an altitude of 5000ft. The plane’s radar shows that it is 8000 ft. from the target site. What is the angle of elevation of the plane measured from the target site to the nearest degree? 5000 ft 8000 ft x h O a

§ 2.5 Using Sine and Cosine Ratios To Calculate Lengths

h a O cos(50) = 5.2 x cos(50) = BC = BC Example #1 Determine the length of BC to the nearest tenth of a centimetre. BC = 3.3 cm What trig ratio uses adjacent and hypotenuse?

h a O sin(55) = Example #2 Determine the length of DE to the nearest tenth of a centimetre. What trig ratio uses opposite and hypotenuse?

Example #3 A surveyor made the measurements shown in the diagram. How could the surveyor determine the distance from the transit to the survey pole to the nearest hundredth of a meter? h a O cos(67.3) = What trig ratio uses adjacent and hypotenuse?

§ 2.6 Applying the Trigonometric Ratios

Example #1 Solve. Give the measures to the nearest tenth. h a o a 2 + b 2 = c = c = c 2 √ 136 = c = c 2 on your calculator: x = tan -1 (10÷6) <x = 59.0° Solve means find the measures of all the sides and angles cm 11.7 = c 59.0° z Angles in a triangle add up to 90° 59.0° + 90° + z = 180° 149.0° + z = 180° ° z = 31.0° 31.0°

Example #2 Solve this triangle. Give the measures to the nearest tenth where necessary. h a o a 2 + b 2 = c = c = c 2 √ = c = c 2 Solve means find the measures of all the sides and angles cm 11.8 = c 11.8 cm z Angles in a triangle add up to 90° 25° + 90° + z = 180° 115° + z = 180° -115° z = 65° 65° 5.0 x tan(65) = Opp Opp = 10.7 cm

Example #3 A helicopter leaves its base, and flies 35km due west to pick up a sick person. It then flies 58km due north to a hospital. a) When the helicopter is at the hospital, how far is it from its base to the nearest kilometre? 35 Km 58 Km x a 2 + b 2 = c = x = x 2 √ 4589 = x = x = x 68 = x

Example #3 A helicopter leaves its base, and flies 35km due west to pick up a sick person. It then flies 58km due north to a hospital. b) When the helicopter is at the hospital, what is the measure of the angle between the path it took due north and the path it will take to return directly to its base? Write the angel to the nearest degree. 35 Km 58 Km 68 Km x h O a on your calculator: x = sin -1 (35÷68) <x = 31° x = cos -1 (58÷68) <x = 31° x = tan -1 (35÷58) <x = 31°

§ 2.7 Solving Problems Involving More than One Right Triangle

h a O h a O 5.7 cm Example #1 Calculate the length of CD to the nearest tenth of a centimetre. x What trig ratio uses opposite and hypotenuse? sin(47) = What trig ratio uses adjacent and hypotenuse? cos(26) = 5.7 x cos(26) = x x = 5.1 cm

Example #2 From the top of a 20-m high building, a surveyor measured the angle of elevation of the top of another building and the angle of depression of the base of the building. The surveyor sketched this plan of her measurements. Determine the height of the taller building to the nearest tenth of a metre. 20 h a o What trig ratio uses opposite and adjacent? tan(15) = 74.6 cm h a o tan(30) = Height = = 63.1 m

Example #3 In the given diagram find. Round your answer to the nearest tenth. x a 2 + b 2 = c = c = c 2 √ 109 = c = c 2 √ 109 h a o What trig ratio uses opposite and adjacent? tan(x) =

Example #4 From the top of a 90-ft. observation tower, a fire ranger observes one fire due west of the tower at an angle of depression of 5 o, and another fire due south of the tower at an angle of depression of 2 o. How far apart are the fires to the nearest foot? h a o What trig ratio uses opposite and adjacent? tan(85) = ft h a o tan(88) = ft a 2 + b 2 = c = c = c 2 √ = c = c ft = c