Chapter 6: Trigonometry 6.2: Trigonometric Applications - PowerPoint PPT Presentation

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Chapter 6: Trigonometry 6.2: Trigonometric Applications

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  1. Chapter 6: Trigonometry6.2: Trigonometric Applications Essential Question:What does SOHCAHTOA mean and how is it used to solve a right triangle problem?

  2. 6.2: Trigonometric Applications • Old stuff will be used in this section • Triangle Sum Theorem • The sum of the measures of the angles in a triangle is 180° • Pythagorean Theorem • In a right triangle with legs a and b and hypotenuse c, a2 + b2 = c2

  3. 6.2: Trigonometric Applications • Finding a side of a Triangle • Find side x in the right triangle below • In this figure, we’re given: An angle (65°) The Hypotenuse (8) A side Adjacent to 65° (x) The sides we’re using are A and H using SOH-CAH-TOA means we use the cosine function 8 65° x

  4. 6.2: Trigonometric Applications • Finding an Angle of a Triangle • Find the measure of the angle θ in the triangle below In this triangle, we’re given all three side lengths, so we can use any of the trigonometric ratios to solve. • SOH • sin θ = 3/5 → sin-1(3/5) = 36.8699° • CAH • cos θ = 4/5 → cos-1(4/5) = 36.8699° • TOA • tan θ = 3/4 → tan-1(3/4) = 36.8699° • All ratios give us the same answer: 36.8699° 4 3 θ 5

  5. 6.2: Trigonometric Applications • Solving a Right Triangle • Solve the right triangle below • The Triangle Sum Theorem helps find θ 75° + θ + 90° = 180°θ = 15° We can use the hypotenuse (17) and the 75° angle to find sides a and b θ 17 a 75° b

  6. 6.2: Trigonometric Applications • Solving a Right Triangle • Solve the right triangle below • The Pythagorean Theorem helps find a a2 + 62 = 122 a2 = 108 a = We can find β by using the cosine functioncos β = 6/12 cos β = 1/2 cos-1(1/2) = β 60° = β We can either find θ by using the sin function or by using The Triangle Sum Theorem θ = 30° θ 12 a β 6

  7. 6.2: Trigonometric Applications • Assignment • Page 429 • Problems 1 – 35, odd problems • Questions where you’re told to not use a calculator can be solved using the chart you copied yesterday.

  8. Chapter 6: Trigonometry6.2: Trigonometric ApplicationsDay 2 Essential Question:What does SOHCAHTOA mean and how is it used to solve a right triangle problem?

  9. 6.2: Trigonometric Applications • Applications • A straight road leads from an ocean beach at a constant upward angle of 3°. How high above sea level is the road at a point 1 mile from the beach? • Answer • If one is not drawn, DRAW A DIAGRAM. • Looking for the side on the right of the triangle, which is the side opposite of 3° • sin 3° = x/52805280 • sin 3° = x276.33 ft = x 5280 ft h 3°

  10. 6.2: Trigonometric Applications • Applications • According to the safety sticker on a 20-foot ladder, the distance from the bottom of the ladder to the base of the wall on which it leans should be one-fourth of the length of the ladder: 5 feet. • How high up the wall will the ladder reach • If the ladder is in this position, what angle does it make with the ground? • Draw a diagram ladder20 ft wallh ft θ ground5 ft

  11. 6.2: Trigonometric Applications • Applications • The wall height can be foundusing the Pythagorean Theorem • 52 + h2 = 202 • h2 = 400 – 25 • h2 = 375 • h = (375)½ ≈ 19.36 ft • We’re given the side adjacent to θ and the hypotenuse, meaning we need to use cosine • cos θ = 5/20 • θ = cos-1(5/20) ≈ 75.5° ladder20 ft wallh ft θ ground5 ft

  12. 6.2: Trigonometric Applications • Angles of Elevation and Depression • Both create right angles from an endpoint. Angles of elevation look up; angles of depression look down. Angle of elevation Angle of depression

  13. 6.2: Trigonometric Applications • Elevation/Depression • A flagpole casts a 60-foot shadow when the angle of elevation of the sun is 35°. Find the height of the flagpole. • Draw a diagram • You’re given a 35° angle • You’re given the side adjacent • You’re looking for the side opposite • You’re using tangent • tan 35° = x / 60 • 60 • tan 35° = x • 42.012 ≈ x flagpoleh ft 35° shadow60 ft

  14. 6.2: Trigonometric Applications • Elevation/Depression (#3: both) • A person on the edge of a canal observes a lamp post on the other side with an angle of elevation of 12° to the top of the lamp post and an angle of depression of 7° to the bottom of the lamp post from eye level. The person’s eye level is 152 cm. • Find the width of the canal. • Find the height of the lamp post. • Draw a diagram top half of lamp post 12° canal 7° 152 cm 152 cm

  15. 6.2: Trigonometric Applications top half of lamp post (y) • Elevation/Depression (#3: both) • The canal is adjacentto the 7° angle • You’re given 152 cm, which is opposite 7° • Use tangent • tan 7° = 152 / x • x = 152 / tan 7° • x = 1237.94 cm • Use the canal measurement to find the top half of the lamp post… again using tangent. • tan 12° = y / 1237.94 • 1237.94 • tan 12° = y • 263.13 cm = y • So the height of the lamp post is 263.13 + 152 = 415.13 cm 12° canal (x) 7° 152 cm

  16. 6.2: Trigonometric Applications • Assignment • Page 431 • Problems 37 – 49, odd problems • Quiz tomorrow • DMS/decimal conversion • Finding 6 trig ratios • Solving right triangles • A word problem or two