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Math 10-3 Chapter 5 – Trigonometry Lesson Plans Day 1: Introduction + Triangle Properties Review Start the class with a Sudoku puzzle. The file: Suduko.doc has a set of 6 that you can print as a set and cut out to use daily. LESSON This lesson is on powerpoint: Trigonometry Day 1 TEACHER.ppt Trigonometry Day 1 STUDENT.ppt We can classify a triangle by…. A) SIDE 1) Scalene (all sides are different lengths) 2) Isosceles (two sides are the same length) 3) Equilateral (all sides are the same length) We can classify a triangle by…. B) ANGLE 1) Right angle (one angle is 90°) 2) Acute (all angles are less than 90°) 3) Obtuse (one angle is greater than 90°) 1 For this course we will be focusing on RIGHT ANGLE triangles and their properties! For ANY triangle, the three angles will ALWAYS add up to total 1800 !!!! Example1: One angle in a right triangle is equal to 350. What is the measure of the third angle? *Draw it!* Hint: A right triangle means one of the angles is = 900 *remember: All angles in a triangle add up to 1800 X + 900 + 350 = 1800 X + 1250 = 1800 X = 1800 - 1250 X = 550 Labeling Triangles Being able to properly label a triangle is a very important skill for trig. -Label the ANGLES (vertexes) with CAPITAL letters -Label the sides with LOWER CASE letters, according to the angle they are opposite -To find the “opposite” side, you always travel “across” from the angle -The 900 is always labeled C -The side opposite the 900 is ALWAYS the hypotenuse; the longest side of a right triangle! - In a right triangle, the two sides that intersect to form a right angle are called legs. - Remember, when two lines meet at a 900 (right) angle, this is called perpendicular! - The side of a triangle that touches an angle is said to be adjacent (beside) that angle. 2 Example2: Label the following right triangle. What side is adjacent to <B? a What side is opposite to <B? b What side is adjacent to <A? b What side is opposite to <A? a What side is the hypotenuse? c What side is opposite <C? c What side is opposite 900? c, OR the hypotenuse! Assignment: M103 Triangle Trivia Assignment.doc You could also play: Triangle Basics Bingo.excl Day 2 – Pythagorean Theorem, Develop the Formula Start the class with a Sudoku puzzle. The file: Suduko.doc has a set of 6 that you can print as a set and cut out to use daily. Play “triangle trivia” with the students using their trivia cards from yesterday. Pit the boys against the girls. 3 LESSON This lesson will be based on discovery learning. Handout the work package: Pythagoras - Discovering the Formula.doc Day 3 – Pythagorean Theorem: Applications Start the class with a Sudoku puzzle. The file: Suduko.doc has a set of 6 that you can print as a set and cut out to use daily. LESSON Discuss with students the formula they discovered yesterday; write some of their ideas on the board. They should have the following ideas: The sum of the areas of the two smaller sides in a right angle triangle equals the area of the longest side. A 3-4-5 triangle is always a right angle triangle; 32 + 42 = 52 a2 + b2 = c2 etc. This lesson is in a word document for the students. Pythagoras NOTES.doc Review: The longest side of a right triangle is called the hypotenuse. We always label this side c. Label the following right triangle: 4 Yesterday, we discovered that there is a relationship between the sides of a right angle triangle. This is called the Pythagorean Theorem Pythagoras came up with a formula to describe this relationship. It is o a2 + b2 = c2 *Remember, when we SQUARE a number, we are multiplying that number by itself. *The opposite of SQUARING a number is the square root The Pythagoras formula is useful for calculating an unknown side of a right angle triangle. Example1: Calculate the hypotenuse of the following right triangle. First, it is always a good idea to label the sides. Use the Pythagoras Formula! *Be sure to show students proper algebra steps* a2 + b2 = c2 42 + 32 = c2 16 + 9 = c2 5 25 = c2 c= 25 c=5 Example 2: Calculate the hypotenuse of the following right triangle. *Label the sides with the students* a2 + b2 = c2 182 + 242 = c2 324 + 576 = c2 900 = c2 c= 900 c = 30 Example 3: Calculate the hypotenuse of the following right triangle to the nearest tenth. *Label the sides with the students* a2 + b2 = c2 52 + 7 2 = c2 25 + 49 = c2 74 = c2 c= 74 c = 8.6 6 Sometimes you may be GIVEN the hypotenuse and a side, and you need to find the other side. This can be a little bit trickier. Example 4: Determine the missing side in the following right triangle. *Label the sides with the students* Discuss how it does not matter which side is a/b a2 + b2 = c2 a2 + 82 = 102 a2 + 64 = 100 a2 = 100 -64 *discuss the algebra step here* a2 = 36 a= 36 a=6 Example 5: Determine the missing side in the following right triangle to the nearest tenth. *Label the sides with the students* a2 + b2 = c2 62 + b2 = 92 36 + b2 = 81 b2 = 81 – 36 7 b2 = 45 b= 45 b = 6.7 Assignment: M103 Pythagoras Applications.doc Day 4 – SINE ratio Start the class with a Sudoku puzzle. The file: Suduko.doc has a set of 6 that you can print as a set and cut out to use daily. LESSON We can use trig functions to solve for unknown sides and angles in 900 triangles. Our calculators must be in degree mode. Trig functions are ratios First, we must properly label our triangle *Remember, the opposite side is across from the angle *The adjacent side is BESIDE the angle The hypotenuse is always across from the 900 and is the LONGEST side First we will work with SINE opposite * Sin( ) OR SOH hypotenuse Ex1. Solve side x to the nearest tenth. 8 opposite Sin( ) hypotenuse Fill in the parts you know x Sin(12) 34 Cross multiply and divide Sin(120) x 34 = x X = 7.1 m Ex2. Solve side x to the nearest tenth. opposite Sin( ) hypotenuse Fill in the parts you know 16 Sin(35) x Cross multiply and divide 9 Sin(350) x = 16 16 X= Sin(35) x = 27.9 cm Assignment: M103 Ch5 Sine.docx Day 5 – Quiz + Sine (Angles) Start the class with the quiz, M103 Ch5 Quiz 1.doc LESSON Sometimes we need to calculate an unknown angle by using the trig ratios To do this, we MUST use the 2nd function on our calculators. This converts the decimal into a degree (angle) Ex1. Sin(X) = 0.3456 Determine the measure of angle X to the nearest degee 2nd sin-1(0.3456) = 200 When given a triangle problem, it is very important to properly label the sides first Ex2. Determine the measure of angle x to the nearest degree 10 opposite *We still use SOH Sin( ) hypotenuse *fill in the parts you know: 7 Sin( x) 15 Sin(x) = 0.4666666…. *Since we are looking for an angle we must use 2nd sin! Sin-1(0.466666….) = 280 Assignment: M103 Sine Angles.doc Day 6 – Cosine Ratio Start the class with a Sudoku puzzle. LESSON opposite Last class we discussed the sine ratio Sin( ) hypotenuse The next trig ratio is COSINE! adjacent * cos( ) OR CAH hypotenuse *This means cosine of an angle is equal to the ratio of the adjacent side divided by the hypotenuse (in a right angle triangle!) Let’s practice labeling a triangle: 11 *Remember: the hypotenuse is always across from the right (900) angle! *Remember: adjacent means beside – it is the side BESIDE the angle! We can use the cosine ratio to solve for sides and angles. Ex1. Solve for the unknown side in the following right triangle, round to the nearest tenth. *show students how to properly label the sides of the triangle* Since we have the hypotenuse, an angle, and we are looking for the adjacent side, we will use the Cosine ratio (CAH). adjacent cos( ) hypotenuse *fill in what you know!* 12 x Cos(160) = 23 *cross multiply and divide!* 23 x Cos(160) = x X = 22.109…. = 22.1 *explain to students how to use their calculators. The may need to type [16 coos x 23 enter] to get the correct answer Double Check! If we are right, then coos(160) = 22.109…/23 Cos(160) = 0.96126…. 22.109…/23 = 0.96126…. They are the same! Ex2. Solve for the unknown side in the following right triangle, round to the nearest tenth. Since we have the adjacent side and the angle, and we are looking for the hypotenuse, we will use the cosine ratio (CAH). adjacent cos( ) hypotenuse 13 *fill in what you know!* 7 Cos(240) = x *cross multiply and divide!* X x Cos(240) = 7 7 X= cos(24) X = 7.662… = 7.7 *remind students about cross multiply, and why it works* Some students may need to use the following key strokes on their calculators [ 7 / (24 cos) enter) Ex3. Calculate the unknown angle in the following right triangle, round to the nearest degree. Since we have the adjacent and hypotenuse, we will use the cosine ratio (CAH) to find the angle. adjacent cos( ) hypotenuse *fill in what you know!* 3 Cos(x) = 8 Cos(x) = 0.375 14 X = cos-1(0.375) = 67.975….. = 680 Assignment: M103 Cosine Ratio.doc Day 7 – Tangent Ratio Start the class with a Sudoku puzzle. LESSON We have already discovered the two trig ratios SINE and COSINE. Today we will work with the last trig ratio, TANGENT! opposite * tan( ) or TOA adjacent *Remember, the opposite SIDE is across from the angle *Remember, adjacent means beside, this side is BESIDE the angle Now we know all three trig ratios to help solve right triangles: SOH CAH TOA *ALWAYS write this on the top of your tests!* Review: Sin(angle) = opp/hyp Cos(angle) = adj/hyp Tan(angle) = opp/adj Practice labeling triangles: 15 Ex1. Calculate the unknown side in the following right triangle. Round to the nearest tenth. Since we have the adjacent side and the angle, and we want to find the opposite side, we will use the Tan ratio (TOA). opposite tan( ) adjacent *Fill in what you know!* x Tan(320) = 17 *cross multiply and divide* 17 x Cos(320) = x X = 14.416…. = 14.4 Ex2. Calculate the unknown side in the following right triangle. Round to the nearest tenth. 16 Since we have the opposite side and the angle, and we want to adjacent side, we will use the Tan ratio (TOA). opposite tan( ) adjacent *fill in what you know!* 102 Tan(160) = x *cross multiply and divide* Tan(160) x X = 102 102 X= Tan(16) X = 355.716…. = 355.7 Ex3. Calculate the unknown angle in the following right triangle. Round to the nearest whole degree. Since we have the opposite and adjacent sides, we will use the Tan ratio (TOA) to calculate the angle. opposite tan( ) adjacent *fill in what you know!* 17 21 Tan(x) = 23 Tan(x) = 0.91304….. X = Tan-1(0.91304….) = 42.39…0 = 420 *Reiterate to students what the “opposite” of Tan(x) is tan-1( Assignment: M103 Tangent Ratio.doc QUIZ NEXT DAY! Day 8 – Quiz on Trig Ratios Start the class with a quick Trig review, being sure to go over: -Labeling right triangles -SOH CAH TOA -How to find an angle when given two sides Quiz – M103 Ch5 Quiz 2.doc After the quiz, play a game (Mario Kart? Poker? Crib? 20 Questions?) Mathematics 10-3 Program of studies Specific Outcomes Analyze puzzles and games that involve spatial reasoning, using problem-solving strategies. Day 9 – Trig Applications Start the class with a Sudoku puzzle or play 20 questions. *This lesson has a notes handout for students* LESSON How do we decide which trig ratio we need to use? 18 *Always label your triangle! This will help us to determine which trig ration (SOH CAH TOA) to use to solve! Ex1. For the triangle above, complete the following: 3 4 3 Tan(A) = Cos(A) = Sin(A) = 4 5 5 Ex2. Brian wants to measure the height of his Christmas tree. He stands 2 m away and measures the angle to the top of his tree to be 650. How tall is the tree? Step 1: Draw a diagram 19 Step 2: Label the sides Step 3: Determine which trig ratio to use (SOH CAH TOA) *Since we have the adjacent side and we want to find the opposite side, we will use TOA (Tangent) Step 4: Solve opposite tan( ) adjacent x Tan(650) = 2 Cross multiply! 2 x Tan (650) = x x = 4.28…m = 4.3 m Clinometers - A clinometer is a device that is used to measure angles of elevation. *The angle of elevation is the angle “looking up” towards an object. *The angle of depression is the angle “looking down” towards an object. Ex2. Lindsay is a surveyor. She needs to calculate the height of a bridge. While standing 15 m from the base of the bridge, she uses a clinometer to measure the angle of elevation to the top of the bridge. She measures 520, which she reads at eye level 1.2 m above the ground. How tall is the bridge? Step 1: Draw it 20 Step 2: Label the triangle Step 3: Determine the trig ratio We have the adjacent side and we want the opposite side, so we will use TOA (Tangent). opposite tan( ) adjacent x Tan(520) = 15 Cross multiply! 15 x Tan(520) = x x = 19.19…. m BUT we have to add the height of Lindsay to this answer! 19.19 m + 1.2 m = 20.39… = 20.4 m Assignment: Triangle Applications.doc Day 10 – Trig Applications Day 2 (Word Problems) Start the class with a Sudoku puzzle. LESSON Word Problems 1) Always read the problem twice. One the second time, underline the important information. Important information includes the numbers involved, and what the question is asking. 2) Sketch a diagram of the problem, 3) Write out the formula you will need to use to solve 21 4) Enter the numbers into the formula and solve. Underline, or put a box around your answer. Ex1: Wayne wants to cut diagonally across a rectangular field with the dimensions of 300 m by 450 m. What is the shortest distance that Wayne will walk? We know that the corners of rectangles are 900. We can use the Pythagoras theorem to solve. a2 + b2 = c2 4502 + 3002 = c2 202500 + 90 000 = c2 292 500 = c2 c= 292500 = 540.8 m Ex2. An airplane has a vertical height of 15 km. If the angle of depression of the plane to the runway is 300, how many km is it to the runway? 22 *remember to label the sides of the triangle! We have the opposite side, we want to solve for the hypotenuse. Which trig ratio do we use? SOH CAH TOA SOH! opposite Sin( ) hypotenuse 15km Sin(300) = x *cross multiply!* X Sin(300) = 15 km 15km X= = 30 km Sin(30) Assignment: Triangle Word Problems.doc Day 11 – Review Project Trig Review Project.doc Day 12 – Review Project (Day 2) – Optional? Day 13 – Ch5 Exam M103 Ch5 Trig Unit Exam.doc 23

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