1. Copyright © 2005 Pearson Education, Inc. Slide 2-1 Solving a Right Triangle To “solve” a right triangle is to find the measures of all the sides and angles of the triangle A right triangle can be solved if either of the following is true:  One side and one acute angle are known  Any two sides are known

2. Copyright © 2005 Pearson Education, Inc. Slide 2-2 Example: Solving a Right Triangle, Given an Angle and a Side Solve right triangle ABC, if A = 42  30' and c = 18.4. How would you find angle B? B = 90  42  30' B = 47  30‘ = 47.5  A C B c = 18.4 42  30'

3. Copyright © 2005 Pearson Education, Inc. Slide 2-3 Example: Solving a Right Triangle Given Two Sides Solve right triangle ABC if a = 11.47 cm and c = 27.82 cm. A C B c = 27.82 a = 11.47

4. Copyright © 2005 Pearson Education, Inc. Slide 2-4 Angles of “Elevation” and “Depression” Angle of Elevation: from point X to point Y (above X) is the acute angle formed by ray XY and a horizontal ray with endpoint X. Angle of Depression: from point X to point Y (below) is the acute angle formed by ray XY and a horizontal ray with endpoint X.

5. Copyright © 2005 Pearson Education, Inc. Slide 2-5 Solving an Applied Trigonometry Problem Step 1Draw a sketch, and label it with the given information. Label the quantity to be found with a variable. Step 2Use the sketch to write an equation relating the given quantities to the variable. Step 3Solve the equation, and check that your answer makes sense.

6. Copyright © 2005 Pearson Education, Inc. Slide 2-6 Example: Application Shelly McCarthy stands 123 ft from the base of a flagpole, and the angle of elevation to the top of the pole is 26 o 40’. If her eyes are 5.30 ft above the ground, find the height of the pole.

7. Copyright © 2005 Pearson Education, Inc. Slide 2-7 Example: Application The length of the shadow of a tree 22.02 m tall is 28.34 m. Find the angle of elevation of the sun. Draw a sketch. The angle of elevation of the sun is 37.85 . 22.02 m 28.34 m B

8. Copyright © 2005 Pearson Education, Inc. Slide 2-8 Describing Direction by Bearing (First Method) Many applications of trigonometry involve “direction” from one point to another Directions may be described in terms of “bearing” and there are two widely used methods The first method designates north as being 0 o and all other directions are described in terms of clockwise rotation from north (in this context the angle is considered “positive”, so east would be bearing 90 o )

9. Copyright © 2005 Pearson Education, Inc. Slide 2-9 Describing Bearing Using First Method Note: All directions can be described as an angle in the interval: [ 0 o, 360 ) Show bearings: 32 o, 164 o, 229 o and 304 o

10. Copyright © 2005 Pearson Education, Inc. Slide 2-10 Hints on Solving Problems Using Bearing Draw a fairly accurate figure showing the situation described in the problem Look at the figure to see if there is a triangular relationship involving the unknown and a trigonometric function Write an equation and solve the problem

11. Copyright © 2005 Pearson Education, Inc. Slide 2-11 Example Radar stations A and B are on an east-west line 3.7 km apart. Station A detects a plane at C on a bearing of 61 o, while station B simultaneously detects the same plane on a bearing of 331 o. Find the distance from A to C.

12. Copyright © 2005 Pearson Education, Inc. Slide 2-12 Describing Direction by Bearing (Second Method) The second method of defining bearing is to indicate degrees of rotation east or west of a north line or east or west of a south line Example: N 30 o W would represent 30 o rotation to the west of a north line Example: S 45 o E would represent 45 o rotation to the east of a south line

13. Copyright © 2005 Pearson Education, Inc. Slide 2-13 Example: Using Bearing An airplane leaves the airport flying at a bearing of N 32  W for 200 miles and lands. How far west of its starting point is the plane? The airplane is approximately 106 miles west of its starting point. e 200

14. Copyright © 2005 Pearson Education, Inc. Slide 2-14 Using Trigonometry to Measure a Distance A method that surveyors use to determine a small distance d between two points P and Q is called the subtense bar method. The subtense bar with length b is centered at Q and situated perpendicular to the line of sight between P and Q. Angle  is measured, then the distance d can be determined.

15. Copyright © 2005 Pearson Education, Inc. Slide 2-15 Example: Using Trigonometry to Measure a Distance Find d when  = and b = 2.0000 cm Let b = 2, change  to decimal degrees.

16. Copyright © 2005 Pearson Education, Inc. Slide 2-16 Example: Solving a Problem Involving Angles of Elevation Sean wants to know the height of a Ferris wheel. He doesn’t know his distance from the base of the wheel, but, from a given point on the ground, he finds the angle of elevation to the top of the Ferris wheel is 42.3 o. He then moves back 75 ft. From the second point, the angle of elevation to the top of the Ferris wheel is 25.4 o. Find the height of the Ferris wheel.

17. Copyright © 2005 Pearson Education, Inc. Slide 2-17 Example: Solving a Problem Involving Angles of Elevation continued The figure shows two unknowns: x and h. Use the two triangles, to write two trig function equations involving the two unknowns: In triangle ABC, In triangle BCD, x C B h DA 75 ft

18. Copyright © 2005 Pearson Education, Inc. Slide 2-18 Example: Solving a Problem Involving Angles of Elevation continued Since each expression equals h, the expressions must be equal to each other. Resulting Equation Distributive Property Factor out x. Get x-terms on one side. Divide by the coefficient of x.

19. Copyright © 2005 Pearson Education, Inc. Slide 2-19 We saw above that Substituting for x. tan 42.3 =.9099299 and tan 25.4 =.4748349. So, tan 42.3 - tan 25.4 =.9099299 -.4748349 =.435095 and  The height of the Ferris wheel is approximately 74 ft. Example: Solving a Problem Involving Angles of Elevation continued

20. Copyright © 2005 Pearson Education, Inc. Slide 2-20 Find of an object in the distance Finding the height of a tree on a mountain.

21. Copyright © 2005 Pearson Education, Inc. Slide 2-21 Find of an object in the distance Finding the height of a tree on a mountain.

22. Copyright © 2005 Pearson Education, Inc. Slide 2-22 Find of an object in the distance Finding the height of a tree on a mountain.

23. Copyright © 2005 Pearson Education, Inc. Slide 2-23 Trigonometry and Bearings Bearing is an acute angle based off the North - South line. N 38º W

24. Copyright © 2005 Pearson Education, Inc. Slide 2-24 A nautical problem A yacht is going 14 knots East for 3 hours, then turns N 42º E for an hour. How far from port is the yacht.

25. Copyright © 2005 Pearson Education, Inc. Slide 2-25 A nautical problem Need to find a hypotenuse of a larger triangle. To find the distance.

26. Copyright © 2005 Pearson Education, Inc. Slide 2-26 A nautical problem The extension helps us find the hypotenuse. We have a few angles and the distance to add.

27. Copyright © 2005 Pearson Education, Inc. Slide 2-27 A nautical problem The extension helps us find the hypotenuse. We have a few angles and the distance to add.

28. Copyright © 2005 Pearson Education, Inc. Slide 2-28 A nautical problem The extension helps us find the hypotenuse. We have a few angles and the distance to add.

29. Copyright © 2005 Pearson Education, Inc. Slide 2-29 A nautical problem The extension helps us find the hypotenuse. We have a few angles and the distance to add.

30. Copyright © 2005 Pearson Education, Inc. Slide 2-30 A nautical problem The extension helps us find the hypotenuse. We have a few angles and the distance to add.

31. Copyright © 2005 Pearson Education, Inc. Slide 2-31 A nautical problem The extension helps us find the hypotenuse. We have a few angles and thedistance to add.

32. Copyright © 2005 Pearson Education, Inc. Slide 2-32 32 Bearings are used to describe direction. Compass bearings have four main directions and four middle directions NE SESW NW E N S W True Bearings are more specific, using a 3-digit angle clockwise from North. Web Practice 1Practice 2 Web Practice 3

33. Copyright © 2005 Pearson Education, Inc. Slide 2-33 33 A Ship A is 17km west of a lighthouse L. A.A.A.A..B.B.B.B.L.L.L.L B Ship B is due south of the lighthouse L. BSE Ship B is SE of ship A. 45 o N 17 km A Calculate distance d from Ship A to ship B. d Cos 45 o = 17 d d = 17 ÷ Cos 45 o ≈ 24 km

34. Copyright © 2005 Pearson Education, Inc. Slide 2-34 34 sails for 20 km on a bearing of 130 o. A Ship sails for 20 km on a bearing of 130 o. 130 o ? How far south is the ship from its starting point? 20 km d 50 o Cos 50 o = d 20 d = 20 x Cos 50 o ≈ 12.9 km

35. Copyright © 2005 Pearson Education, Inc. Slide 2-35 35 Mary walks 3.4 km east, then 1.3 km south. 3.4 km 1.3 km θoθoθoθo Find Mary’s bearing from her starting position. Tan θ o = 1.3 3.4 θo =θo = Tan -1 ( ) 1.3 3.4 90 o ≈ 21 o Bearing = 90 o + 21 o = 111 o