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Grade 11 Essential Mathematics Trigonometry Grade 11 Essential Mathematics Unit 5: Trigonometry Page 1 Grade 11 Essential Mathematics Trigonometry Unit 5: Trigonometry Introduction: This unit deal with Pythagorean Theorem, and the trigonometric ratios of sine, cosine, and tangent. You will be using these to solve word problems as well. Trigonometry is based on the relationship between the measure of the angles and the lengths of the sides of a right angle triangle. These skills are necessary in occupations such as carpentry, aviation and astronomy. You will also learn how to solve 2D and 3D triangles. Assessment: o o o o o o o Lesson 1 Assignment: Problem Solving using Pythagorean Theorem Lesson 2 Assignment: SOH CAH TOA Lesson 3 Assignment: Word Problems using Trigonometric Ratios Lesson 4 Assignment: Angles of Elevation and Declination Lesson 5 Assignment: Two Triangle Problems Putting It Together β Trigonometry Lesson 6 Assignment: Solving 3-D Triangle Problems Unit 5: Test Trigonometry Page 2 Grade 11 Essential Mathematics Trigonometry LESSON 1: PYTHAGOREAN THEOREM The Pythagorean Theorem is for right angle triangle only. Pythagorean Theorem: c a Hypotenuse π2 + π 2 = π 2 b The hypotenuse is the side that is always opposite (across from) the 90° angle. This is always side C in the Pythagorean Theorem. Example: Determine the length of the missing for the following triangle. Solution: x 3 π₯ 2 = 32 + 42 π₯ 2 = 9 + 16 π₯ 2 = 25 π₯ = β25 = 5 4 Example: Determine the length of the missing side for each of the following triangle. Solution: 102 = π₯ 2 + 82 100 = π₯ 2 + 64 100 β 64 = π₯ 2 80 = π₯ 2 π₯ = β80 = 8.9 Example: Determine the length of the missing side for each of the following triangle. Solution: 152 = π₯ 2 + 72 225 = π₯ 2 + 49 225 β 49 = π₯ 2 360 = π₯ 2 π₯ = β360 = 18.9 10 x 8 15 x 7 Page 3 Grade 11 Essential Mathematics Trigonometry Word Problems using the Pythagorean Theorem Example: Scott wants to swim across a river that is 400m wide. He plans to swim directly across the river but ends up 100m downstream because of the current. How far did he actually swim? Solution: Step 1: Draw a diagram for the right angle triangle: 400 m x 100 m Step 2: Label the sides of the triangle. Pay attention to where the hypotenuse is!!! 400 m x HYP 100 m π₯ 2 = 4002 + 1002 π₯ 2 = 160000 + 10000 π₯ 2 = 170000 π₯ = β170000 = 412.3π Page 4 Grade 11 Essential Mathematics Trigonometry Example: To get from point A to point B you must avoid walking through a building. To avoid the building, you walk 14m south and 25m east. How many metres would you have saved had the building not been there? Solution: Must walk 14 + 25 = 39π x 14 m 142 + 252 = π₯ 2 196 + 625 = π₯ 2 821 = π₯ 2 π₯ = β821 = 28.65π 25 m If was able to walk through the building would only have to walk 28.65 m. So would save 39 β 28.65 = 10.35π Example: The foot of a 6m ladder is placed 2m from the base of a building. How far up the building does that ladder reach? Solution: 6m x π₯ 2 + 22 = 62 π₯ 2 + 4 = 36 π₯ 2 = 36 β 4 π₯ 2 = 32 π₯ = β32 = 5.7π 2m Page 5 Grade 11 Essential Mathematics Trigonometry Example: Calculate x to the nearest tenth of a metre. 12 m 7m x 4m Solution: π¦ 2 + 72 = 122 π¦ 2 + 49 = 144 π₯ 2 = 144 β 49 π₯ 2 = 95 π₯ = β95 = 9.75π π₯ 2 + 42 = 9.752 π¦ 2 + 16 = 95 π₯ 2 = 95 β 16 π₯ 2 = 79 π₯ = β79 = 8.89π Page 6 Grade 11 Essential Mathematics Trigonometry Curriculum Outcomes: 11E4.Develop a spatial sense related to triangles. Lesson 1 Assignment: Problem Solving using Pythagorean Theorem See your teacher for Lesson 1 Assignment Page 7 Grade 11 Essential Mathematics Trigonometry LESSON 2: TRIGONOMETRY The three trig functions are SIN, COS, and TAN SOH CAH TOA Stands for: sin π = πππ βπ¦π cos π = πππ βπ¦π tan π = πππ πππ When we are using the trigonometric functions, we need to label the sides of the given right angle triangle so that we know what number goes where in our formulas. The three sides are: 1. Hypotenuse: is the side of the triangle that is always opposite the 90° angle 2. Opposite: is the side that is opposite (across from) the given angle, we never use the 90° angle to find the opposite side 3. Adjacent: is the side that is adjacent (next to) the given angle we never use the 90° angle to find the adjacent side Example: Label the sides of the following triangle. ΞΈ Solution: The side that is across from the 90° angle is the Hypotenuse. The side that is across from the angle π± is the Opposite and the side that is next to the given angle π± is the Adjacent side. Page 8 Grade 11 Essential Mathematics Trigonometry Using the calculator: First, we need to make sure that our calculator is set in Degrees. To do this, check the screen on the calculator, it should have a D, DEG, or a DRG. If it has any of the following we need to change it to Degrees: R, RAD, G, GRAD. To change the settings on the calculator look for a button that had DRG on it. Press it until you see D, DRG or DEG on the screen of your calculator. This will make sure that your calculator is in Degrees. It is important that we do this since all of the angles in this unit are measured in degrees, if your calculator is set in something different all of your answers will be incorrect. Depending on your calculator you will either enter the trigonometric function first or the number first. Example: π ππ 45 = 0.7071 Example: πππ 73 = 0.2924 Example: π‘ππ 54 = 1.3764 We can also use the functions to determine the size of the angle. These are called the Inverse Trig Functions. These are found on the calculator and not as their own buttons. To use these functions we must press the INV, Shift, or 2nd button on the calculator Example: πππ ο± = 0.7 Solution: π ππβ1 π πππ β1 π π‘ππβ1 π π = cos β1 0.7 π = 45° Example: π ππ ο± = 0.15 Solution: π = sinβ1 0.15 π = 8.6° Example: π‘ππ ο± = 6.2 Solution: π = tanβ1 6.2 π = 81° Page 9 Grade 11 Essential Mathematics Trigonometry Trigonometric Ratios Example: Determine the length of the missing side. 15.2 7 320 x x 0 45 Solution: sin 45 = π₯ sin 32 = 7 π₯ 0.7071 = 7 π₯ 15.2 π₯ πΆπππ π ππ’ππ‘ππππ¦ π‘π π πππ£π πππ π₯ 0.5299 = 15.2 (0.7071)(7) = π₯ (0.5299)(15.2) = π₯ 4.95 = π₯ 8.05 = π₯ Example: Solve for the hypotenuse. x x 220 4.6 7 0 45 Solution: sin 45 = 0.7071 = 4.6 π₯ 4.6 π₯ (0.7071)(π₯) = 4.6 4.6 sin 22 = π₯ = 0.7071 = 6.5 7 π₯ 7 πΆπππ π ππ’ππ‘ππππ¦ 0.3746 = π₯ π·ππ£πππ π‘π π πππ£π πππ π₯ (0.3746)(π₯) = 7 7 π₯ = 0.3746 = 18.7 Page 10 Grade 11 Essential Mathematics Trigonometry Example: Solve for the missing angle. 21 15 ΞΈ 8 16 ΞΈ Solution: sin π = 8 sin π = 15 16 21 sin π = 0.5333 sin π = 0.7619 π = sinβ1 0.5333 = 32.2° π = sinβ1 0.7691 = 50.3° Example: Determine the length of the missing side. 19 7.5 320 x 450 x Solution: cos 45 = π₯ cos 32 = 7.5 π₯ 0.7071 = 7.5 πΆπππ π ππ’ππ‘ππππ¦ π‘π π πππ£π πππ π₯ π₯ 19 π₯ 0.848 = 19 (0.7071)(7.5) = π₯ (0.848)(19) = π₯ 5.3 = π₯ 16.11 = π₯ Page 11 Grade 11 Essential Mathematics Trigonometry Example: Solve for the hypotenuse. x x 220 7.9 450 7.6 Solution: cos 45 = 0.7071 = 7.6 cos 22 = π₯ 7.6 π₯ (0.7071)(π₯) = 7.6 7.9 π₯ 7.9 πΆπππ π ππ’ππ‘ππππ¦ 0.9272 = π·ππ£πππ π‘π π πππ£π πππ π₯ (0.9272)(π₯) = 7.9 7.6 π₯ 7.9 π₯ = 0.7071 = 5.4 π₯ = 0.9272 = 8.5 Example: Solve for the missing angle. ΞΈ 21 15 8 ΞΈ 16 Solution: cos π = 8 15 cos π = 16 21 cos π = 0.5333 cos π = 0.7619 π = cos β1 0.5333 = 57.7° π = cos β1 0.7691 = 40.4° Page 12 Grade 11 Essential Mathematics Trigonometry Example: Determine the length of the missing side. 320 x x 10.2 0 45 7 Solution: tan 45 = π₯ tan 32 = 7 π₯ 1 = 7.5 π₯ 10.2 π₯ πΆπππ π ππ’ππ‘ππππ¦ π‘π π πππ£π πππ π₯ 0.6249 = 19 (1)(7.5) = π₯ (0.6249)(19) = π₯ 7.5 = π₯ 11.8 = π₯ Example:. Solve for the missing side. 220 16 7 450 Solution: tan 45 = 1= x 16 16 π₯ 16 1 tan 22 = π₯ (1)(π₯) = 16 π₯= x = 16 7 π₯ 7 πΆπππ π ππ’ππ‘ππππ¦ 0.4040 = π₯ π·ππ£πππ π‘π π πππ£π πππ π₯ (0.4040)(π₯) = 7 7 π₯ = 0.4040 = 17.3 Page 13 Grade 11 Essential Mathematics Trigonometry Example: Solve for the missing angle. ΞΈ 8 16 12 ΞΈ 5 Solution: tan π = 8 15 tan π = 16 21 tan π = 0.5333 tan π = 0.7619 π = tanβ1 0.5333 = 86.2° π = tanβ1 0.7691 = 37.3° Page 14 Grade 11 Essential Mathematics Trigonometry Curriculum Outcomes: 10E2.TG.3. Solve problems that require the manipulation and application of primary trigonometric ratios 10E2.TG.2.Demonstrate an understanding of primary trigonometric ratios (sine, cosine, tangent) Lesson 2 Assignment: SOH CAH TOA See your teacher for Lesson 2 Assignment Page 15 Grade 11 Essential Mathematics Trigonometry LESSON 3: USING THE TRIG RATIOS TO SOLVE WORD PROBLEMS Example: Suppose a kite handle has 15ft of kite string. If the wind picks up and the kite string makes an angle with the ground of 56°, what is the height of the kite? Solution: π₯ sin 56 = 15 π₯ 15 ft 0.829 = 15 x ft (0.829)(15) = π₯ 12.4 ππ‘ = π₯ 56° πβπ πππ‘π ππ 12.4 ππππ‘ πππ π‘βπ ππππ’ππ Example: A wire supports a tower and forms an angle of 57° with the ground. The wire is attached to the ground at a point that is 8.5m away from the base of the tower. a. at what height is the wire attached to the tower? b. how long is the wire? Solution: π‘ππ57 = π₯ 8.5 π₯ 1.5399 = 8.5 ym (1.5399)(8.5) = π₯ xm 13.1π = π₯ 57° 8.5 cos 57 = π¦ 0.5446 = π¦= π‘βπ π€πππ ππ ππ‘π‘ππβππ 13.1π πππ π‘βπ ππππ’ππ 8.5 m 8.5 π¦ 8.5 = 15.6π 0.5446 π‘βπ π€πππ ππ 15.6π ππππ Page 16 Grade 11 Essential Mathematics Trigonometry Example: A 6.1m ladder leans against a wall. The angle formed by the ladder and the wall is 71°. a. how far is the base of the ladder from the wall? b. how far up the wall does the ladder reach? Solution: π₯ sin 71 = 6.1 71° π₯ 0.9455 = 6.1 (0.9455)(6.1) = π₯ 6.1m 5.8 π = π₯ ym πβπ ππππππ ππ 5.8 π ππππ π‘βπ π€πππ cos 71 = π¦ 6.1 xm (0.3256)(6.1) = π¦ π¦ = 1.98 π π‘βπ ππππππ ππππβππ 2 π π’π π‘βπ π€πππ Example: A truck travels 6km up a mountain road. The change in height is 1.25km. What is the angle of the road? Solution: sin π = 1.25 km 1.25 6 sin π = 0.2083 6 km π = sinβ1 0.2083 π± π = 12° Page 17 Grade 11 Essential Mathematics Trigonometry Curriculum Outcomes: 10E2.TG.3. Solve problems that require the manipulation and application of primary trigonometric ratios 10E2.TG.2.Demonstrate an understanding of primary trigonometric ratios (sine, cosine, tangent) Lesson 3 Assignment: Word Problems using Trigonometric Ratios See your teacher for Lesson 3 Assignment Page 18 Grade 11 Essential Mathematics Trigonometry LESSON 4: ANGLE OF ELEVATION AND DECLINATION Definition: 1. Angle of Elevation: is the angle formed up from the horizontal 2. Angle of Declination: is the angle formed down from the horizontal Example: From a point 8m from the base of a building, you measure the angle up to the top of the building from eye level and find that it is 50°. If you are 1.2m tall, how tall is the building? Solution: tan 50ο° = x 1.19 = y 50ο° π₯ 8 π₯ 8 (1.19)(8) = π₯ 8m 1.2m π₯ = 9.5π ππ π‘βπ βπππβπ‘ ππ π‘βπ π‘πππ ππ 9.5 + 1.2 = 10.7 π π‘πππ Page 19 Grade 11 Essential Mathematics Trigonometry Example: The highest point on a cliff is 90m above the shore. From the top of the cliff, a surveyor measures the angle of declination to a boat in the lake to be 42°. How far away from shore is the boat? Solution: x tan 42ο° = 42ο° 90m 0.9 = π₯= 90 π₯ 90 π₯ 90 0.9 π₯ = 100π x Example: From a point 5m from the base of a tree, you measure the angle of elevation using a 1.5m tall instrument to be 39°. How tall is the tree? Solution: tan 39ο° = 0.81 = y X π¦ 5 π¦ 5 π¦ = (5)(0.81) = 4.05π 39ο° 1.5m 5m ππ π‘βπ βπππβπ‘ ππ π‘βπ π‘πππ ππ π = 4.05 + 1.5 = 5.55π π‘πππ. Page 20 Grade 11 Essential Mathematics Trigonometry Example: A 5m tall lighthouse sits at the top of a 30m cliff and the top of the lighthouse is 35m above sea level. The angle of depression to a fishing boat is 24°. The angle of depression to a second boat past the fishing boat is 16°. How far apart are the two boats? Solution: 24ο° y 16ο° 35m 35m 35m x π‘π ππππ βππ€ πππ πππππ‘ π‘βπ ππππ‘π πππ ππππ πππβππ‘βππ π€π ππππ π‘π πππππ’πππ‘π: π = π β π πΉπππ π‘ π€π ππππ π‘π ππππ€ βππ€ πππ ππ€ππ¦ ππππ‘ 1 ππ ππππ π‘βπ πππππ 35 tan 24ο° = π§ 0.445 = π§= 35 π§ 35 = 78.7π 0.445 π΅πππ, ππ πππππππππ πππ π πππππππ ππππ π ππ ππππ πππ πππππ π= πππ§ ππο° = ππ π π. πππ = ππ π ππ = πππ. ππ π. πππ πΏ = πππ. π β ππ. π = ππ. ππ ππ πππ πππππ πππ ππ. ππ πππππ Page 21 Grade 11 Essential Mathematics Trigonometry Curriculum Outcomes: 11E4.Develop a spatial sense related to triangles. Lesson 4 Assignment: Angles of Elevation and Declination See your teacher for Lesson 4 Assignment Page 22 Grade 11 Essential Mathematics Trigonometry LESSON 5: SOLVING TWO-TRIANGLE PROBLEMS Example: Nolan works as a tour guide in Rankin Inlet and is often asked to take pictures for tourists standing beside the statue of an inuksuk. If Nolanβs friend Michael is 187cm tall and is standing beside the statue, how tall is the inuksuk? Nolan has measured two angles, one to the top of Michaelβs head is 26° and the other to the top of the inuksuk is 50°. Solution: tan 26 = 187 0.4877 = x 187 187 π¦ (π¦)(0.4877) = 187 (π¦)(0.4877) 50° π¦ 0.4877 187 = 0.4877 187 π¦ = 0.4877 = 383.4ππ 26° y tan 50 = ππ πππππ ππ π π‘ππππππ 383.4ππ ππ€ππ¦ ππππ π‘βπ π π‘ππ‘π’π π₯ 383.4 1.1918 = π₯ 383.4 (1.1918)(383.4) = π₯ 456.9ππ = π₯ ππ π‘βπ π π‘ππ‘π’π ππ 456.9ππ π‘πππ. Page 23 Grade 11 Essential Mathematics Trigonometry Example: A camp instructor leads a group of students on a canoe trip across a lake. They leave Half Moon Bay (H) and paddle 7.8km to Jarvis Bay (J). They then turn 69° and paddle 5.1km to a beach. This beach, B, is 7.6km across the lake from Half Moon Bay. At what angle should they turn in order to where they started? Solution: Even though the diagram looks as though we can cut line segment BJ in half, we cannot assume that this is actually true. Our diagram might not be accurate so we cannot make this assumption. sin 69 = π¦ 7.8 0.9336 = π¦ 7.8 (0.9336)(7.8) = π¦ 7.28ππ = π₯ sin π = 7.28 7.6 sin π = 0.9579 π = sinβ1 0.9579 = 73° πβππ¦ π βππ’ππ π‘π’ππ ππ‘ ππ πππππ ππ 73° Page 24 Grade 11 Essential Mathematics Trigonometry Example: Two office towers are 50m apart. From the top of the shorter tower to the top of the taller towers a worker measures the angle of elevation to be 25° and the angle of depression to the base of the taller tower to be 35°. Determine the height of each of the towers? Solution: Remember that both angle of elevation and depression are measured off of a Horizontal line. We must include this horizontal line from the spot where the angles are measured. The observer is standing on the roof of the shorter office building and it is from this point that the angles are measured so here is where the horizontal line is drawn. Angle of Elevation of 25° is UP from the horizontal and angle of depression of 35° is Down from the horizontal. We want to know the height of both of the towers. We can calculate the height of the shorter tower right away: π₯ 50 π₯ 0.7002 = 50 (0.7002)(50) = π₯ tan 35 = y 25° 35° π₯ = 35π 50m z πβπ π βπππ‘ππ ππ’ππππππ ππ 35π x x 50m πππ§ ππ = π ππ π. ππππ = π ππ (π. ππππ)(ππ) = π ππ. ππ = π π = π + π = ππ + ππ. π = ππ. ππ πππ ππππ ππ πππππ πππ ππ ππ. ππ Page 25 Grade 11 Essential Mathematics Trigonometry Curriculum Outcomes: 11E4.TG.1. Solve problems that involve two and three right triangles Lesson 5 Assignment: Two Triangle Problems See your teacher for Lesson 5 Assignment Page 26 Grade 11 Essential Mathematics Trigonometry See your teacher for Assignment βPutting It Together β Trigonometryβ Page 27 Grade 11 Essential Mathematics Trigonometry LESSON 6: SOLVING 3-D TRIANGLE PROBLEMS Examples: Determine the length of AY in the diagram below. Solution: πΌπ πππππ π‘π πππππ’πππ‘π π‘βπ πππππ‘β ππ π‘βπ πππππ πππ‘π‘ππ ππππ π€π ππππ π‘π πππ‘ππππππ π‘βπ πππππ‘β ππ π πππππ π€π πππβ² π‘ βππ£π πππ¦ ππππππ π€π βππ£π π‘π π’π π ππ¦π‘βπππππππ πβπ π2 + π 2 = π 2 122 + 92 = π₯ 2 144 + 81 = π₯ 2 225 = π₯ 2 π₯ = β225 = 15ππ 82 + 15 = π 2 64 + 225 = π 2 289 = π 2 π = β289 = 17ππ Page 28 Grade 11 Essential Mathematics Trigonometry Example: Carl and Devon are rock climbing instructors. They need to determine the height of the peak they will be climbing. From where Carl is standing it is 51° up to the top of the cliff they will be climbing. Devon is 105m away from Carl. The angle between base camp and Carl is measured by Devon to be 78°. How high is the cliff that they will be climbing? Solution: x 78° 105m π€πππ‘ π‘π ππππ€ π‘βπ βπππβπ‘ ππ π‘βπ πππππβ π΄π΅ ππ ππππ π΄π΅ π€π ππππ π‘ ππππ π‘π ππππ€ π‘βπ πππππ‘β ππ π π₯ tan 78 = 105 π₯ 105 (4.7046)(105) = π₯ 4.7046 = 493.99 π = π₯ ππ πππ πππ€ π’π π π₯ = 494 π π‘π πππππ’πππ‘π π‘βπ βπππβπ‘ ππ π‘βπ πππππ π¦ tan 51 = 494 π¦ 1.2349 = 494 (1.2349)(494) = π¦ π¦ = 610 π Page 29 Grade 11 Essential Mathematics Trigonometry Example: Anne works in a lighthouse that is 160 feet above sea level. She sees a boat south east from the lighthouse and a second boat that is south west of the lighthouse. She measures the angle of depression to the first boat measures to be 24° and the angle of depression to the second boat to be 38°. How far apart are the two boats? Solution: 160ft a b x πππ‘π ππππ ππ‘ πππβ ππππ‘ π ππππππ‘πππ¦. π ππππππ π€π πππ πππ£ππ π‘βπ πππππ ππ π·πππππ π πππ π π π€π ππππ π‘π ππππ€ π‘βπ βππππ§πππ‘ππ ππππ ππππ π€βπππ π‘βπ πππππ ππ ππππ π’πππ ππ‘ π‘βπ π‘ππ ππ π‘βπ πππβπ‘βππ’π π. a 24° 160ft 160ft πππ§ ππ = πππ π π. ππππ = πππ π (π)(π. ππππ) = πππ a π= πππ = πππ. πππ π. ππππ ππ ππππ‘ ππ’ππππ 1 ππ 359.4 ππππ‘ ππππ π‘βπ πππ π ππ π‘βπ πππβπ‘βππ’π π Page 30 Grade 11 Essential Mathematics Trigonometry b 38° 160ft 160ft πππ§ ππ = πππ π π. ππππ = πππ π (π)(π. ππππ) = πππ b πππ π = π.ππππ = πππ. πππ ππ ππππ‘ ππ’ππππ 2 ππ 204.8 ππππ‘ ππππ π‘βπ πππ π ππ π‘βπ πππβπ‘βππ’π π π2 + π 2 = π 2 359.42 + 204.82 = π₯ 2 x b 129168.36 + 41943.04 = π₯ 2 171107.57 = π₯ 2 a π₯ = β171107.57 = 413.65 ππ‘ ππ π‘βπ ππππ‘π πππ 413.65 ππππ‘ πππππ‘ Page 31 Grade 11 Essential Mathematics Trigonometry Curriculum Outcomes: 11E4.TG.1. Solve problems that involve two and three right triangles. Lesson 6 Assignment: Solving 3-D Triangle Problems See your teacher for Lesson 6 Assignment Page 32
