M103 Ch5 Trig Lesson Plans

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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



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