1 / 25

Chapter 3

Chapter 3. Acute Triangle Trigonometry. 3.1 – Exploring Side -Angle Relationships in Acute Triangles. Chapter 3: Acute Triangle Trigonometry. Trigonometry. What are the three trigonometric functions?. Sine (sin), Cosine ( cos ), and Tangent (tan). Remember:. SOH CAH TOA. Hypotenuse

wilona
Download Presentation

Chapter 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 3 Acute Triangle Trigonometry

  2. 3.1 – Exploring Side-Angle Relationships in Acute Triangles Chapter 3: Acute Triangle Trigonometry

  3. Trigonometry What are the three trigonometric functions? Sine (sin), Cosine (cos), and Tangent (tan) Remember: SOH CAH TOA Hypotenuse Opposite sin Adjacent Opposite tan Hypotenuse Adjacent cos

  4. Example • Solve for height, h: • sin(46.5) = h/51.2cm (sin=opp/hyp) • h = 51.2 × sin(46.5) • h = 37.1 cm • Or: • sin(83.3) = h/37.3cm • h = 37.3 × sin(83.3) • h = 37.1 cm • What’s the relationship between 37.3 cm and 46.5°? • How about 51.2 cm and 83.3°? • What would have happened if we had put the h going from vertex B down to its opposite side? What could we say about 39.7 cm and 50.4°? So: 37.3 × sin(83.3) = 51.2 × sin(46.5) Or:

  5. The Sine Law Law of Sines: OR

  6. Pg. 131, # 1-5 Independent Practice

  7. 3.2 – Proving and Applying the Sine Law Chapter 3: Acute Triangle

  8. Last time we learned the Sine Law: Proof: Draw a triangle like the last, with another perpendicular line from b to vertex B, called h2. h2 • h = b x sin(C) = c x sin(B) • b = c . • sin(B) sin(C) • h2 = c x sin(A) = a x sin(C) • c = a . • sin(C) sin(A) sin(C) = h/b sin(B) = h/c sin(A) = h2/c sin(C) = h2/a Remember! sin = opp/hyp

  9. Example A triangle has angles measuring 80° and 55°. The side opposite the 80° angle is 12.0 m in length. Determine the length of the side opposite the 55°angle to the nearest tenth of a metre. Draw a diagram: What will the third angle be? What’s the total sum of angles in a triangle? Sine Law: Which side are we looking for? To the nearest tenth of a metre, the side opposite the 55° angle is 10.0 m long.

  10. Example Toby uses chains attached to hooks on the ceiling and a winch to lift engines at his father’s garage. The chains, the winch, and the ceiling are arranged as shown. Toby solved the triangle using the sine law to determine the angle that each chain makes with the ceiling to the nearest degree. He claims that θ=40° and α=54°. Is he correct? Explain, and make any necessary corrections. So, if θ=51°, how can we tell what α is? Remember, the sum of the angles must be 180°. θ+α+86° = 180° 51°+α+86° = 180° α=180°-51°-86°=43° So, Toby was incorrect, in actuality θ=51° and α=43°.

  11. Example The captain of a small boat is delivering supplies to two lighthouses, as shown. His compass indicates that the lighthouse to his left is located at N30°W and the lighthouse to his right is located at N50°E. Determine the compass direction he must follow when he leaves lighthouse B lighthouse A. We need to draw a line pointing East from vertex B. O Since the new line goes West to East, it must be perpendicular to the line going South to North… so the angle is 90°. If we know that one angle in ΔBCO is 90° and another is 50°, how can we figure out the third angle? ∠NBO = 47.6° - 40° = 7.6° To find the direction on the compass, we must subtract our answer from 90°. So, he will head N82°W from lighthouse B. The angle B is 47.612°, but we want the direction he’d go from lighthouse B. ∠CBO = 180° - 50° - 90° = 40°

  12. Pg. 138-141, # 2-15, 18 Independent Practice

  13. 3.3 – Proving and Applying the Cosine Law Chapter 3: Acute Triangle Trigonometry

  14. Cosine Law In any acute triangle, a2= b2 +c2 – 2bccosA b2= a2 + c2 – 2accosB c2 = a2 + b2 – 2abcosC

  15. Proof of the Cosine Law In any acute triangle, a2 = b2 + c2 – 2bccosA b2= a2 + c2 – 2accosB c2 = a2 + b2 – 2abcosC • Proof: • h2 = c2 – x2 • h2 = b2 – y2 •  c2 – x2 = b2 – y2 • c2 = x2 + b2 – y2 • c2 = (a – y)2 + b2 – y2 (since x = a – y) • c2 = a2 – 2ay + y2 + b2 – y2 • c2 = a2 + b2 – 2ay cosC = y/b (cos = adj/hyp) bcosC = y c2 = a2+ b2 – 2ay c2= a2 + b2 – 2ab cosC The other parts can be proven similarly.

  16. Example Determine the length of CB to the nearest metre. a2 = b2 + c2 – 2bccosA a2= 322+ 402– 2(32)(40)cos(58°) a2 = 1024 + 1600– 2560cos(58°) a2 = 2624 – 2560cos(58°) a2 = 1267.406 a = 35.600… Why can’t you use the Sine Law? CB is 36 m.

  17. Example A three-pointed star is made up of an equilateral triangle and three congruent isosceles triangles. Determine the length of each side of the equilateral triangle in this three-pointed star. Round the length to the nearest centimetre. First, label the diagram! x2 = z2 + y2 – 2zycosX x2 = 602 + 602 – 2(60)(60)cos(20°) x2 = 3600 + 3600 – 6765.786… x2= 434.213 x = 20.837… Each side of the equilateral triangle has a length of 21 cm.

  18. PG. 150 – 153, #1, 4, 5, 7, 8, 10, 11, 13, 15. Independent Practice

  19. 3.4 – Solving Problems Using Acute Triangles Chapter 3: Acute Triangle Trigonometry

  20. Example Two security cameras in a museum must be adjusted to monitor a new display of fossils. The cameras are mounted 6 m above the floor, directly across from each other on opposite walls. The walls are 12 m apart. The fossils are displayed in cases made of wood and glass. The top of the display is 1.5 m above the floor. The distance from the camera on the left to the centre of the top of the display is 4.8 m. Both cameras must aim at the centre of the top of the display. Determine the angles of depression, to the nearest degree, for each camera. sinθ=4.5/4.8 θ=sin-1(4.5/4.8) θ=69.635…° Draw a diagram:

  21. Example continued… z2 = d2 + b2 – 2dbcosA z2 = 122+ 4.82– 2(12)(4.8) cos(69.635) z2 = 126.952… z = 11.267… sinα= 4.5/11.267 α= sin-1(4.5/11.267) α= 23.539° Camera A must be adjusted to an angle of 70° and camera B must be adjusted to an angle of 24°.

  22. Example Brendan and Diana plan to climb the cliff at Dry Island Buffalo Jump, Alberta. They need to know the height of the climb before they start. Brendan stands at point B, as shown in the diagram. He uses a clinometer to determine ∠ABC, the angle of elevation to the top of the cliff. Then he estimates ∠CBD, the angle between the base of the cliff, himself, and Diana, who is standing at point D. Diana estimates ∠CDB, the angle between the base of the cliff, herself, and Brendan. Determine the height of the cliff to the nearest metre. In ΔDBC, ∠BCD = 180° - 60° - 50° ∠BCD= 70°

  23. Example continued… The height of the cliff is 196 m.

  24. Pg. 161 – 164, #1-14, 16. Independent Practice

  25. Activity Cut out the seven similar triangles from the handout, and use all seven of them to form a single square. Show me once you succeed. If the hypotenuse of the greatest triangle is 10 units long, what is the area of the square?

More Related